Basic Laws. Associative Law for Addition and Multiplication. Commutative Law for Addition and Multiplication. Distributive Law : a*(b c) = a*b a*c

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1 CHAPTER 1F Fraction 1 F1 The Basics Whole numbers 1,,,,.. are called natural numbers and they are used to track whole items. If x and y are two natural numbers and x can be divided by y, we say x is a multiple of y or y is a factor of x. Hence 1 is a multiple of (or ), and (or ) is a factor of 1. Note that a number x is always the multiple of itself and the number 1. A prime number is a natural number that can only be divided by 1 and itself. Its factors consist of 1 and itself only. Hence,,, 7, 11, 1, are prime numbers. 1 is not a prime number because 1 can be divided by or. The number 1 is not a prime number. Example A. List the factors and the multiples of 1. The factors of 1: 1,,,, 6, and 1. The multiples of 1: 1=1x1, =x1, 6=x1, 8, 60, etc To factor a number x means to write x as a product, i.e. write x as a*b for some a and b. For example 1 can be factored as 1 = *6 = * = **. We say the factorization is complete if all the factors are prime numbers. Hence 1=** is factored completely, *6 is not complete because 6 is not a prime number. Exponents To simplify writing repetitive multiplication, we write for *, we write for **, we write for *** and so on. We write x*x*x *x as x N where N is the number of x s multiplied to itself. N is called the exponent, or the power of x. Example B. Calculate. = * = 9 (note that * = 6) = ** = 6 (note that * = 1) Numbers that are factored completely may be written using the exponential notation. Hence, factored completely, 1 = ** = *, 00 = 8* = **** = *. Each number has a unique form when its completely factored into prime factors and arranged in order. This is analogous to chemistry where each chemical has a unique chemical composition of basic elements such as H 0 for water. Larger numbers may be factored using a vertical format. Example C. Factor 1 completely. (Vertical format) 1 Gather all the prime numbers at the end of the branches we have = ***** =. Note that we obtain the same answer regardless how we factor at each step. 1

2 Associative Law for Addition and Multiplication ( a + b ) + c = a + ( b + c ) ( a * b ) * c = a * ( b * c ) Basic Laws Of the four arithmetic operations +,, *, and, +, and * behave nicer than or. We often take advantage of the this nice behavior of the addition and multiplication operations to get short cuts. However, be sure you don t mistaken apply these short cuts to and. Note: We are to perform the operations inside the ( ) first. For example, (1 + ) + = + = 6, which is the same as 1 + ( + ) = 1 + = 6. Subtraction and division are not associative. For example, ( ) 1 = 0 is different from ( 1 ) =. Commutative Law for Addition and Multiplication a + b = b + a a * b = b * a For example, * = * = 1. Subtraction and division don't satisfy commutative laws. For example: 1 1. From the above laws, we get the following important facts. When adding, the order of addition doesn t matter. Hence to add a list numbers, it's easier to add the ones that add to multiples of 10 first. ExampleD = = 91 When multiplying, the order of multiplication doesn t matter. When multiplying many numbers, always multiply them in pairs first. This is useful for raising exponents. Example E. 6 = ***** = 9 * 9 * 9 = 81 * 9 = 79 Distributive Law : a*(b c) = a*b a*c Example F. *( + ) = *7 = or by distribute law, *( + ) = * + * = =

3 Basic Laws Exercise A. Do the following problems two ways. * Add the following by summing the multiples of 10 first. * Add by adding in the order. to find the correct answer Basic Laws B. Calculate List the all the factors and the first multiples of the following numbers. 6, 9, 10, 1, 16,, 0, 6,, 6, Factor completely and arrange the factors from smallest to the largest in the exponential notation:, 8, 1, 16, 18,, 7,, 6,, 8, 6, 60, 6, 7, 7, 81, 10. C. Multiply in two ways to find the correct answer. 1. * * *. 6 * * *. 6 * 1 * *. 7 * * *. 6 * 7 * * 6. 9 * * * 7. * * * * 8. * * * * * 9. * * * * * 0. * * * 1 * 8 * F Fractions p Fractions are numbers of the form q (or p/q) where p, q 0 are whole numbers. Fractions are numbers that measure parts of whole items. Suppose a pizza is cut into 6 equal slices and we have of them, the fraction that represents this quantity is. 6 6 The top number is the number of parts that we have and it is called the numerator. The bottom number is the number of equal parts in the division and it is called the denominator.

4 Whole numbers can be viewed as fractions with denominator 1. x 0 Thus = and x =. The fraction = 0, where x x However, x does not have any meaning, it is undefined. 0 The Ultimate No-No of Mathematics: The denominator (bottom) of a fraction can't be 0. (It's undefined if the denominator is 0.) Fractions that represents the same quantity are called equivalent fractions. 1 are equivalent fractions. = = 6 = 8 The fraction with the smallest denominator of all the equivalent fractions is called the reduced fraction. 1 is the reduced one in the above list. Factor Cancellation Rule Given a fraction a, then a b b = a b / / c c that is, if the numerator and denominator are divided by the same quantity c, the result will be an equivalent fraction. In other words, a common factor of the numerator and the a*c a denominator may be canceled as 1, i.e. = a*c 1 = b*c b*c b. (Often we omit writing the 1 s after the cancellation.) To reduce a fraction, we keep divide the top and bottom by common numbers until no more division is possible. What's left is the reduced version. 78 Example A: Reduce the fraction. 78 = 78/ 9 = 9/ = 1 / 7 7/ 9. or divide both by 6 in one step. One common mistake in cancellation is to cancel a common number that is part of an addition (or subtraction) in the numerator or denominator. A participant in a sum or a difference is called a term. The in the expression + is a term (of the expression). The is in the expression * is called a factor. Terms may not be cancelled. Only factors may be canceled. = = + 1 +!? = 1 This is addition. Can t cancel! * 1 * = 1 Yes Improper Fractions and Mixed Numbers A fraction whose numerator is the same or more than its denominator (e.g. ) is said to be improper. We may put an improper fraction into mixed form by division.

5 Improper Fractions and Mixed Numbers Example B. Put into mixed form. = with remainder. Hence, = + =. We may put a mixed number into improper fraction by doing the reverse via multiplication. Example C: Put into improper form. = * + = Fractions and Mixed Numbers Exercise. A. Reduce the following fractions , 1, 9, 18,, 6, 8, 108 B. Convert the following improper fractions into mixed numbers then convert the mixed numbers back to the improper form F Multiplication and Division of Fractions Rule for Multiplication of Fractions To multiply fractions, multiply the numerators and multiply the denominators, but always cancel as much as possible first then multiply. a c d = a*c b * b*d Example A. Multiply by reducing first * 1 a. = 8 * 8 * * = = * *8*9*10 b. 8 * 9 * * 11 = = 7 8*9*10*11 11 a c Can't do this for addition and subtraction, i.e. d = a c b ± ± b ± d

6 The fractional multiplications a b d d a * or * are important. b Often in these problems the denominator b can be cancelled against d = d. 1 Example B: Multiply by cancelling first. a. 6 * 18 = * 6 = 1 b * = * 11 = 16 The often used phrases " (fraction) of.." are translated to multiplications correspond to this kind of problems. Example C: a. What is of $108? 6 The statement translates into * 108 = * 6 = 7 $. b. A bag of mixed candy contains 8 pieces of chocolate, caramel and lemon drops. 1/ of them are chocolate, 1/ of them are caramel. How many pieces of each are there? What fraction of the candies are lemon drops? For chocolate, ¼ of 8 is 1 1 * 8 = 1, so there are 1 pieces of chocolate candies. 16 For caramel, 1/ of 8 is 1 * 8 = 16, so there are 16 pieces of caramel candies. The rest = 0 are lemon drops. The fraction of the lemon drops is 0 8 = 0/ 8/ = 1 c. A class has x students, ¾ of them are girls, how many girls are there? It translates into multiplication as * x. Reciprocal and Division of Fractions The reciprocal (multiplicative inverse) of a is b. b a So the reciprocal of is, the reciprocal of is 1, the reciprocal of 1 and the reciprocal of x is 1 is, x. Two Important Facts About Reciprocals I. The product of x with its reciprocal is * = 1, * = 1, x * x = 1, II. Dividing by x is the same as multiplying by its reciprocal 1 x. For example, 10 is the same as 10 1 *, both yield. Rule for Division of Fractions To divide by a fraction x, restate it as multiplying by the reciprocal 1/x, that is, a c a d b d = b * c reciprocate = a*d b*c 6

7 Example D: Divide the following fractions. a. b = * 8 = = * = d. 1 6 = * 6 1 = 0 Example E: We have ¾ cups of sugar. A cookie recipe calls for 1/16 cup of sugar for each cookie. How many cookies can we make? We can make 16 1 = 16 * 1 = * = 1 cookies. Multiplication and Division of Fractions Exercise A. Do the following problems like the example above. Do them mentally. 1. a. * b. * c. 7 * d. 9 *. a. * 6 b. * 6 c. * 6 d. * 6. a. *6 b. * 6 c. * 6 6 d. 9 * 6. a. *8 b. *8 c. 7 *8 d. *8. a. *9 9 b. *9 c. *1 d. 9 *1 6. a. *1 b. *1 c. 7 6 *1 d. *1 7. a. *18 b. *18 c. 7 6 *18 d. 7 9 *18 8. a. * b. * c. 7 * d. * 6 9. a. 7 * 1 b. 11 * 6 c. 7 *6 d. *6 6 7

8 10. a. *6 b. *6 9 c. 7 *6 1 d. *6 Exercise. B. 11. In a class of people, / of them are boys, how many boys are there? 1. In a class of 8 people, 1/ of them are boys, how many girls are there? 1. In a class of 60 people, / of them are not boys, how many boys are there? 1. In a class of 7 people, /6 of them are not girls, how many boys are there? 1. In a class of 6 people, /7 of them are not boys, how many girls are there? 16. In a class of 60 people, 1/ of them are girls, how many are not girls? 17. In a class of 60 people, / of them are not girls, how are not boys? 18. In a class of 108 people, /9 of them are girls, how many are not boys? A mixed bag of candies has 7 pieces of colored candies, 1/8 of them are red, 1/ of them are green, ½ of them are blue and the rest are yellow. 19. How many green ones are there? 0. How many are blue? 1. How many are not yellow? 0. How many are not blue and not green? 1. In a group of 108 people, /9 of them adults (aged 18 or over), 1/ of them are teens (aged from 1 to 17) and the rest are children. Of the adults / are females, / of the teens are males and 1/ of the children are girls. Complete the following table. Male Female Total Adults Teens Children. How many females are there and what is the fraction of the females to entire group?. How many are not male adults and what is the fraction of them to entire group? 8

9 1 F LCM and LCD Definition of LCM The least common multiple (LCM) of two or more numbers is the least number that is the multiple of all of these numbers. Example A. Find the LCM of and 6. The multiples of are, 8, 1, 16, 0,, The multiples of 6 are 6, 1, 18,, 0, The smallest of the common multiples is 1, so LCM{, 6 } = 1. We may improve the above listing-method for finding the LCM. Given two or more numbers, we start with the largest number, list its multiples in order until we find the one that can divide all the other numbers. Example B. Find the LCM of 8, 9, and 1. The largest number is 1 and the multiples of 1 are 1,, 6, 8, 60, 7, 8 The first number that is also a multiple of 8 and 9 is 7. Hence LCM{8, 9, 1} = 7. But when the LCM is large, the listing method is cumbersome. It's easier to find the LCM by constructing it instead. To construct the LCM: a. Factor each number completely b. For each prime factor, take the highest power appearing in the factorizations. The LCM is their product. Example C. Construct the LCM of {8, 1, 18}. Factor each number completely, 8 = 1 = * 18 = * From the factorization select the highest degree of each prime factor:,,, then LCM{8, 1, 18} = * * = 8*9* = 60. The LCM of the denominators of a list of fractions is called the least common denominator (LCD). Following is an application of the LCM. Example D. From one pizza, Joe wants 1/, Mary wants 1/ and Chuck wants 1/6. How many equal slices should we cut the pizza into and how many slices should each person take? What is the fractional amount of the pizza they want in total? In picture: Joe Mary Chuck We find the LCM of 1/, 1/, 1/6 by searching. The multiples of 6 are 6, 1, 18,, Since 1 is also the multiple of and, then 1 is the LCM. Hence we should cut it into 1 slices and Joe gets 1 1* = slices Mary gets 1 1* = slices Chuck gets 1 1* = slices 6 9 In total, that is + + = 9 slices, or 1 = of the pizza. 9

10 Your Turn: From one pizza, Joe wants /8 of it, Mary wants 1/6 of it and Chuck wants /1 of it. How many equal slices should we cut the pizza and how many slices should each person take? In the above example, we found that 1 is the same 1. The following theorem tells us how to convert the denominator of a fraction to a fraction with a different denominator. Multiplier Theorem: To convert the fraction a into a fraction with denominator d, b the new numerator is a * d. b 9 Example D: Convert 16 to a fraction with denominator 8. The new denominator is 8, then the new numerator is 9 Hence 9 8* = 7. = LCM and LCD Exercise A. Find the LCM. 1. a.{6, 8} b. {6, 9} c. {, } d. {, 10}. a.{, 6, 8} b. {, 6, 9} c. {,, } d. {, 6, 10}. a.{6, 8, 9} b. {6, 9, 10} c. {, 9, 10} d. {6, 8, 10}. a.{, 8, 1} b. {8, 9, 1} c. {6, 9, 1}. a.{6, 8, 1} b. {8, 9, 1} c. {6, 9, 16} 6. a.{8, 1, 1} b. { 9, 1, 1} c. { 9, 1, 16} 7. a.{8, 1, 18} b. {8, 1, 0} c. { 1, 1, 16} 8. a.{8, 1, 1, 18} b. {8, 1, 16, 0} 9. a.{8, 1, 18, 0} b. {9, 16, 0, } B. Convert the fractions to fractions with the given denominators. 10. Convert 7 to denominator 1.,, 6, 11. Convert 1 to denominator. 6,, 6, 8 1. Convert to denominator 6. 1,, 9, 6 1. Convert to denominator , 1,, 1 10

11 1 F Addition and Subtraction of Fractions Suppose a pizza is cut into equal slices and Joe takes one slice or ¼ of the pizza, Mary takes two slices or / of the pizza, altogether they take 1 + = of the entire pizza. In picture: 1 + = Addition and Subtraction of Fractions With the Same Denominator To add or subtract fractions of the same denominator, keep the same denominator, add or subtract the numerators a b d d = a d ± b,then simplify the result. Suppose a pizza is cut into equal slices and Joe takes one slice or ¼ of the pizza, Mary takes two slices or / of the pizza, altogether they take 1 + = of the entire pizza. In picture: 1 + = Addition and Subtraction of Fractions With the Same Denominator To add or subtract fractions of the same denominator, keep the same denominator, add or subtract the numerators a b d d = a d ± b,then simplify the result. Example A: a = 18 = 18/6 1 = 1 1/6 = b. 8 + = = = Fractions with different denominators can t be added directly since the size of the fractions don t match. For example 6 + To add them, first find the LCD of ½ and 1/, which is 6. We then cut each pizza into 6 slices. Both fractions may be converted to have the denominator 6. Specifically, 6 = 6 1 = 6 1 = 6 Hence, = = 6 11

12 Addition and Subtraction of Fractions Example A: a = 18 1 = 18/6 1 = 1/6 = b. 8 + = 8 + = = Fractions with different denominators can t be added directly since the size of the fractions don t match. For example 6 + To add them, first find the LCD of ½ and 1/, which is 6. We then cut each pizza into 6 slices. Both fractions may be converted to have the denominator 6. Specifically, 1 = 6 1 = 6 6 = 6 Hence, = = 6 We need to convert fractions of different denominators to a common denominator in order to add or subtract them. The easiest common denominator to use is the LCD, the least common denominator. We list the steps below. Addition and Subtraction of Fractions With the Different Denominator 1. Find their LCD. Convert all the different-denominator-fractions to the have the LCD as the denominator.. Add and subtract the adjusted fractions then simplify the result. Example B: a Step 1: To find the LCD, list the multiples of 8 which are 8, 16,,.. we see that the LCD is. Step : Convert each fraction to have as the denominator. For, the new numerator is * 6 6 = 0, hence 6 = 0 For, the new numerator is * = 9, hence = 9 Step : Add the converted fractions = = 9 7 b The LCD is 8. Convert: = * 1 = 8 so 1 = = * 8 = 0 so 8 = 0 8 = * 8 16 = 7 so 16 = 7 8 1

13 We may use the following multiplier-method to add or subtract fractions to reduce the amount of writing. This method is based on the fact that if we multiply a quantity x by a, then divide by a, we get back x. For example, * / = 10/ =, * 8 / 8 = /8 =. Multiplier Method (for adding and subtracting fractions) To add or subtract fractions, multiply the problem by the LCD (expand it distributive using law), then divide by the LCD. Example C: a The LCD is. Multiply the problem by, then divide by. ( ) * / = (* + *) / = 9/ = 9 b The LCD is 8. Multiply the problem by 8, expand the multiplication, divide the result by ( ) * / 8 = (*7 + 6* *9) / 8 = ( ) / 8 = 8 1 Addition and Subtraction of Fractions Exercise A. Calculate and simplify the answers B. Calculate by the Multiplier Method and simplify the answers

14 C. Addition and Subtraction of Fractions Addition and Subtraction of Fractions F6 Some Facts About Divisibility We start out with a simple mathematics procedure that is often used in real live. It s called the digit sum. Just as its name suggests, we sum all the digits in a number. The digit sum of 1 is 1 + =, is the same as the digit sum of 1, 111, or To find the digit sum of add add Hence the digit sum of is. If we keep adding the digits, the sums eventually become a single digit sum the digit root. The digit root of is 9. 1

15 A bag contains 8 pieces of chocolate, can they be evenly divided by 1 kids? In mathematics, we ask is 8 divisible by 1? for short. How about,9,876,,1 pieces of chocolate with 18 kids? Is,9,876,,1 divisible by 18? One simple application of the digit sum is to check if a number may be divided completely by numbers such as 9, 1, or 18. I. The Digit Sum Test for Divisibility by and 9. If the digit sum or digit root of a number may be divided by (or 9) then the number itself maybe divided by (or 9). For example 1, 111, and all have digit sums that may be divided by, therefore all of them may be divided by evenly. However only , whose digit sum is 9, may be divided evenly by 9. Example A. Identify which of the following numbers are divisible by and which are divisible by 9 by inspection. a. b. 600 c. 61 d We refer the above digit sum check for and 9 as test I. We continue with test II and III. II. The Test for Divisibility by,, and 8 A number is divisible by if its last digit is even. A number is divisible by if its last digits is divisible by you may ignore all the digits in front of them. Hence **** is divisible by but **** is not. A number is divisible by 8 if its last digits is divisible by 8. Hence ****880 is divisible by 8, but ****80 is not. III. The Test for Divisibility by. A number is divisible by if its last digit is. From the above checks, we get the following checks for important numbers such as 6, 1, 1, 18, 6, etc.. The idea is to do multiple checks on any given numbers. The Multiple Checks Principle. If a number passes two different of tests I, II, or III, then it s divisible by the product of the numbers tested. The number 10 is divisible by -via the digit sum test. It is divisible by because the last digit is even. Hence 10 is divisible by * = 6 (10 = 6 *17). Example B. A bag contains 8 pieces of chocolate, can they be evenly divided by 1 kids? How about,9,876,,1 pieces of chocolate with 18 kids? For 8 its digit sum is 1 so it s divisible by (but not 9). Its last two digits are 8 which is divisible by. Hence 8 is divisible by * or 1. For,9,876,,10 for 18, since it's divisible by, we only have to test divisibility for 9. Instead of actually find the digit sum, let s cross out the digits sum to multiple of 9. We see that it s divisible by 9, hence it s divisible by 18. 1

16 Some Facts About Divisibility Exercise. Check each for divisibility by or 6 by inspection Check each for divisibility by or 8 by inspection Which numbers in problems 1 10 are divisible by 9? 1. Which numbers in problems 1 10 are divisible by 1? 1. Which numbers in problems 1 10 are divisible by 18? 1 F7 Cross Multiplication In this section we look at the useful procedure of cross multiplcation. Cross Multiplication Many procedures with two fractions utilize the operation of cross multiplication as shown below. Take the denominators and multiply them diagonally across. What we get are two numbers. a b ad c d bc Make sure that the denominators cross over and up so the numerators stay put. Do not cross downward as shown here. a b c d bc ad Here are some operations where we may cross multiply. Rephrasing Fractional Ratios If a cookie recipe calls for cups of sugar and cups of flour, we say the ratio of sugar to flour is to, and it s written as : for sugar : flour. Or we said the ratio of flour : sugar is :. For most people a recipe that calls for the fractional ratio of / cup sugar to / cup of flour is confusing. It s better to cross multiply to rewrite this ratio in whole numbers. Example A. rewrite a recipe that calls for the fractional ratio of / cup sugar to / cup of flour into ratio of whole numbers. Write / cup of sugar as and / cup of flour as S F. We have the ratio S : F cross multiply we ve 9S : 8F. Hence in integers, the ratio is 9 : 8 for sugar : flour. Remark: A ratio such as 8 : should be simplified to : 1. 16

17 Cross Multiplication Test for Comparing Two Fractions When comparing two fractions to see which is larger and which is smaller. Cross multiply them, the side with the larger product corresponds to the larger fraction. In particular, if the cross multiplication products are the same then the fraction are the same. 9 Hence cross multiply 1 we get = so 9 = 1 Cross multiply we get 8 less more Hence is less than 8. (Which is more or 11? Do it by inspection.) Cross Multiplication Test for Comparing Two Fractions When comparing two fractions to see which is larger and which is smaller. Cross multiply them, the side with the larger product corresponds to the larger fraction. In particular, if the cross multiplication products are the same then the fraction are the same. 9 Hence cross multiply 1 we get = so 9 = 1 Cross multiply we get 8 less more Hence is less than 8. (Which is more or 11? Do it by inspection.) Cross Multiplication for Addition or Subtraction We may cross multiply to add or subtract two fractions with the product of the denominators as the common denominator. a b c ± d = ad ± bc bd Afterwards we reduce if necessary for the simplified answer. Example B. Calculate a. 6 = * 6* 6* = 7 0 b. 9 1 = *1 9* = 1 = 9* Cross multiplication is the same as clearing the two denominators by using their product. We may use their LCM to clear the denominators but the LCM requires extra work. 17

18 Cross Multiplication for Addition or Subtraction We may cross multiply to add or subtract two fractions with the product of the denominators as the common denominator. a b c ± d = ad ± bc bd Afterwards we reduce if necessary for the simplified answer. Example B. Calculate a. 6 = * 6* 6* = 7 0 b. 9 1 = *1 9* = 1 = 9* Cross multiplication is the same as clearing the two denominators by using their product. We may use their LCM to clear the denominators but the LCM requires extra work. Cross Multiplication Exercise A. Restate the following ratios in integers :. 1 1 :. :. :. 1 : : 7. : 8. : In a market, ¾ of an apple may be traded with ½ a pear. Restate this using integers. B. Determine which fraction is more and which is less. 10., ,,, , ,,, C. Use cross multiplication to combine the fractions

19 CHAPTER 1B Signed Numbers 1 S1 Signed Numbers We track the directions of measurements by giving them positive (+) or negative (-) sign. Signed measurements represent amounts of increases versus decreases, surpluses versus deficiencies, credits versus debits and so on. Numbers with signs are called signed numbers. Example A: a. We deposited $00 into a bank account then withdrew $0 from the account, write the transactions using signed numbers. How much is left in the account? We use + for deposit or having surplus in the account and for withdraw or debit from the account, then the transactions may be listed as: +00, 0. (In this section we will use red color for negative numbers for emphasis.) The amount left in the account is $0. This is a surplus so its +0. We write the entire transactions as = +0. b. We deposited $00 into the account then withdrew $600 from the account, write the transactions using signed numbers. How much is left in the account? The transactions may be listed as: +00, 600. The account is short by $00. This is a deficiency so its 00. We write the entire transactions as = 00 c. We deposited $00 in a bank account, later deposited another $00, then withdrew $0, then deposited $0, then withdrew $600, make a list of the transactions using signed numbers. How much is left? The transactions are listed as +00, +00, 0, +0, 600. To find the final balance in the account, we note that the total of the deposits is +80 and the total of the withdrawals is 90, so the account is short of $100, or there is 100 left in the account. We write these transactions as = 100. The above operation of totaling two or more signed numbers into a single signed number is called the combining operation. Example B: = +00, = = +00, = 00 A number written without a sign is treated as a positive number. Therefore, is the same as and both combined to 00. In order to state precisely the rules for combining signed numbers, we introduce the notion of absolute values. In example A of the bank account, if we are only interested in the amount of the transactions but not the type of transactions, this amount is called the absolute value. The absolute value of a number x is written as x. Hence, 00 = 00, -0 = 0, -600 = 600, etc.. 19

20 Rules for Combining Signed Numbers I. To combine two or more numbers of the same signs, keep the sign, take the sum of the absolute values of the numbers = +00, = 00 II. To combine two numbers of different signs, keep the sign of the number with larger absolute value, take the difference of the absolute values of the numbers = +00, = 00 There are different ways to combine multiple signed numbers, we may combine them from left to right. Example C: = = 10 To combine many numbers, an alternative way is to do it as in example A of the bank transactions. That is, we combined all the positive ones (deposits) first, then combine all the negative ones (withdrawals), then combine the two results. Exercise A. Combine Signed Numbers B. Combine by moving the positive numbers to the front first. Combine the positive numbers, the negative numbers separately then then combine the two results Try to get the same answer for #0 by combining two numbers at a time without separating the positive numbers from the negative numbers. 0

21 1 S Addition and Subtraction of Signed Numbers Addition of Signed Numbers Adding signed numbers is the same as combining the numbers. Rule for Addition of Signed Numbers: To add two signed numbers, remove the parenthesis and combine the numbers, that is, a (+b) = a + b, a + ( b) = a b Example A. Remove parentheses then combine. a. (+) remove ( ) = + = 9 b. 7 () remove ( ) = 7 + = c. ( 6) remove ( ) = 6 = d. ( 8) remove ( ) = 8 = 1 e. () + ( ) + ( 8) + 6 remove ( ) s = = = 1 17 = Subtraction of Signed Numbers For subtraction of signed numbers, we need the notion of opposite numbers. The numbers x and x are said to be the opposite or the negative of each other. The opposite of x is x. The opposite of x is ( x) = x. So the opposite of 6 is 6, the opposite of 1 is ( 1) = 1. Note that the opposite of a negative number is positive. Rule for Subtraction of Signed Numbers: To subtract a signed number x, remove the parenthesis and combine with the opposite of x, that is, opposite a (+b) = a b opposite a ( b) = a + b Example B. Remove parentheses then combine. a. (+) remove ( ), change to opposite = = 1 b. ( 7) remove ( ), change to opposite = + 7 = 10 c. 1 ( ) remove ( ), change to opposite = 1 + = 7 Summary for removing parentheses for addition and subtraction of signed numbers: For addition: a (+b) = a + b a + ( b) = a b For subtraction: a (+b) = a b a ( b) = a + b Example C. Remove the parentheses, then combine. a. 6 ( 8) ( ) (9) remove ( ) = = = 10 1 = 1

22 1 S Addition and Subtraction of Signed Numbers b. ( ) ( 8) () ( 9 ) remove the ( ) s = = = = 8 The ( ), [ ], or { } are grouping symbols. Each set of symbol must contain both the left-hand and right-hand parts. Each set encloses calculations that are to be done within the symbol. Example D. a. 6 (8 9) do the calculation inside the ( ) = 6 ( 1) remove parentheses = = b. ( 6 8) 9 do the calculation inside the ( ) = ( 1) 9 remove parentheses = 1 9 = If there is a set of grouping symbol inside another set of grouping symbol, the inner set is to be calculated first. Example E. [ 6 (8 + 9)] do the calculation inside the ( ) = [ 6 (17)] remove parentheses = [ 6 17] do the calculation inside the [ ] = [ ] = + = 1 S Addition and Subtraction of Signed Numbers Exercise A. Drop the parentheses and write the expressions in a simpler form, then combine (+). 7 ( ). 7 (). 7 ( ). 7 () ( ) ( ) ( 1) 9. 8 ( 1) 9 B. Drop the parentheses after the subtraction (don t forget to switch the sign) and write the expressions in a simpler form, then combine (+) 1. 7 ( ) 1. 7 () 1. 7 ( ) 1. 7 () ( ) ( ) (+1) ( 1) ( 1) 9 + ( ) 0. 8 ( 1) ( 19) (+)

23 Addition and Subtraction of Signed Numbers C. Drop the parentheses after the arithmetic operations and write the expressions in a simpler form, then combine (+) + ( ). 7 ( ) + (). 7 ( ) + ( ). 8 + ( 6) ( ) (). 17 ( 6) + ( 1) ( 1) 1 ( ) 18 ( 1) ( ) + + ( 16) ( 1) ( ) 10 D. Work the problems starting from the inside. Make sure that you copy the ( ) s, [ ] s correctly [+ + ( )] 0. [7 ( )] + () 1. 7 [( ) + ( )]. 8 + [( 6) ( )]. 8 + ( 6) [( ) ]. [ 1 1] [( ) 18] ( 1) 6. [8 ( 1)] [9 + ] [7 + ( ) 10]. [17 ( )] [( 16) 11] 1 S Multiplication and Division of Signed Numbers Rule for Multiplication of Signed Numbers To multiple two signed numbers, we multiply their absolute values and use the following rules for the sign of the product. + * + = * = + ; + * = * + = ; Two numbers with the same sign multiplied yield a positive product. Two numbers with opposite signs multiplied yield a negative product. Example A. a. * () = * ( ) = 0 b. * () = * ( ) = 0 In algebra, multiplication operation are not always written down explicitly. Instead we use the following rules to identify multiplication operations. If there is no operation indicated between two quantities, the operation between them is multiplication. Hence xy means x * y. If there is no operation indicated between a set of ( ) and a quantity, the operation between them is multiplication. Hence x(a + b) = x * (a + b ). If there is no operation indicated between two sets of ( ) s, the operation between them is multiplication. Hence (x + y)(a + b) = (x + y) * (a + b) However, if there is a + or sign between the ( ) and a quantity, then the operation is to combine. Hence (+) = (+) =1, but + () = () + = 8, and ( ) = ( )( ) =, but ( ) = () = 10. When multiplying many signed numbers at once, find the sign of the product first by the following rules.

24 The General Sign Rule (for * and /) If there are even number of negative numbers in the multiplication, the product is positive. If there are odd number of negative numbers in the multiplication, the product is negative. Example B. a. 1( ) ( 1) = three negative numbers, so the product is negative b. ( ) = ( )( )( )( ) = 16 four negative numbers, so the product is positive Fact: A quantity raised to an even power i.e. x even is always positive (or 0). General Rule for the Sign of a Product If there are even number of negative numbers in the multiplication, the product is positive. If there are odd number of negative numbers in the multiplication, the product is negative. Example B. a. 1( ) ( 1) = three negative numbers, so the product is negative b. ( ) = ( )( )( )( ) = 16 four negative numbers, so the product is positive Fact: A quantity raised to an even power i.e. x even is always positive (or 0). Multiplication and Division of Signed Numbers Exercise A. Calculate the following expressions. Make sure that you interpret the operations correctly. 1.. ( ). (). ( ). ( ) 6. ( )( ) 7. ( ) ( ) 8. ( ) ( ) B.Multiply. Determine the sign first. 9. ( ) 10. ( )( ) 11. ( 1)( )( ) 1. ( )( ) 1. ( )( )( ) 1. ( )( )( )( ) 1. ( 1)( )( )( )( ) 16. ( 1)()( 1)( ) C. Simplify. Determine the sign and cancel first ()( 6). ( 18)( 6). ( 9)(6) 8 9 (1)( ). (1)( ) ( 8)( 10) 7. ( )( ) ( )( 1)( ). ( 1)( 9) ( 7)(1) 8. ( )()( )( 6) ( )()( )6( 7) 6. ( )( 6)( 1) ()( )( )

25 1 S Order of Operations If we have two $-bill and two $10-bills, we have the total of () + (10) = 0 dollars. To get the correct answer 0, we multiply the and the and multiply the and the10 first, then we add the products 10 and 0. If I have three $10-bills and you have four $10-bills, we have + = 7 $10-bills, and we have a total of ( + )10 = 70 $. In this case, we group the + in the ( ) to indicate that we are to add them first, then multiply the sum to 10. This motivates us to set the rules for the order of operations. Order of Operations (excluding raising power) Given an arithmetic expression, we perform the operations in the following order. 1st. Do the operations within grouping symbols, starting with the innermost grouping symbol. nd. Do multiplications and divisions (from left to right). rd. Do additions and subtractions (from left to right). Example A. a. ( 8) + () = + 1 = 17 b. + ( + ) = + (7) = + 1 = c. 9 [7 (6 + 1)] = 9 [7 (7)] = 9 [7 1] = 9 [ 1 ] = = 7 (Don t perform + or 9 in the above problems!!) Do now. Don t do the part that you shouldn t do! ( + 1). 10 ( ). + [ + (1 + )]. [ + ( 9)] Ans: a. 18 b. 18 c.. 1 Exponents We write x*x*x *x as x N where N is the number of copies of x s multiplied to itself. N is called the exponent, or the power of x, and x is called the base. The base is the quantity immediately beneath the exponent, hence b means *b = *b*b*b. If we want multiply b to itself three times, i.e. b to the third power, we write it as (b) which is (b)*(b)*(b) =8b.

26 Example B. (Exponential Notation) a. Expand ( ) and simplify the answer. The base is ( ). Hence ( ) is ( )( ) = 9 b. Expand The base of the nd power is. Hence means (*) = 9 c. Expand (*) and simplify the answer. The base for the nd power is (*). Hence(*) is (*)(*) = (6)(6) = 6 d. Expand * and simplify the answer. The base for the nd power is. Hence * means ** = 1 e. Expand ( y) and simplify the answer. ( y) = ( y)( y)( y) (the product of three negatives number is negative) = ()()()(y)(y)(y) = 7y From part b above, we see that the power is to be carried out before multiplication. Below is the complete rules of order of operations. Order of Operations (PEMDAS) 1 st. (Parenthesis) Do the operations within grouping symbols, starting with the innermost one. nd. (Exponents) Do the exponentiation rd. (Multiplication and Division) Do multiplications and divisions in order from left to right. th. (Addition and Subtraction) Do additions and subtractions in order from left to right. e. Expand ( y) and simplify the answer. ( y) = ( y)( y)( y) (the product of three negatives number is negative) = ()()()(y)(y)(y) = 7y From part b above, we see that the power is to be carried out before multiplication. Below is the complete rules of order of operations. Order of Operations (PEMDAS) 1 st. (Parenthesis) Do the operations within grouping symbols, starting with the innermost one. nd. (Exponents) Do the exponentiation rd. (Multiplication and Division) Do multiplications and divisions in order from left to right. th. (Addition and Subtraction) Do additions and subtractions in order from left to right. 6

27 Example C. Order of Operations a. = 9 = 16 b. ( ) = () = c. * + (*) = *9 + (6) = = 18 d. ( 6) = 9 ( ) = 9 (9) = 9 = Order of Operations Order of Operations Exercise A. Calculate the following expressions. Make sure that you interpret the operations correctly. 1. ( ). (). (). ( ) +. +( )(+) 6. + ( )(+) B. Make sure that you don t do the ± too early () 8. (6) ( 9) ( ) 11. 6(7 8) 1. ( ) [1 ( )] [ + 6(7 8)](+) [1 ( + )] 16. 6[ 6(7 8)] [1 ( )] [ + 6(7 8)] 19. (1 + )[1 ( + )] 0. ( 6)[ 6(7 8)] 1. 1 ( )( ). ( )( 6) ( 7)( 8) C. Calculate. Make sure apply the powers to the correct bases.. ( ) and ( ) and. ( ) and 6. ( ) and 7. * 8. (*) D.Make sure that you apply the powers to the correct bases. 9. () () ( ) + ( ) 1 1. () + (). ( 1) + ( 1). ( ) ( ) 1. () () + () 1. ( 1) ( 1) + ( 1) 1 6. ( ) ( ) ( ) E. Calculate. 7. (6 + ) ( + ) 0. ( ) + () (7 + )(7 ). ( ). ( + )( ). 6. ( ) ( + * + ) ( + )( * + ) 9. () ()() 0. () (1)( ) 1. ( ) ( )(). ( ) ( 1)( ). 7 ( ). 8 6 ( ). ( ) ( 8) 6. ( 7) ( ) ( ) ( ) ( 6) 7

28 1 S Variables and Evaluation In mathematics we use symbols such as x, y and z to represent numbers. These symbols are called variables because their values change depending on the situation. We use variables and mathematics operations to make expressions which are calculation procedures. For example, if an apple cost $ and x represents the number of apples, then x is the expression for the cost for x apples. Suppose we have 6 apples, set x = 6 in the expression x, we obtain (6) = 1 for the total cost. The value 6 for x is called input (value). The answer 1 is called the output. This process of replacing the variables with input value(s) and find the output is called evaluation. Each variable can represent one specific measurement only. Suppose we need an expression for the total cost of apples and pears and x represents the number of apples, we must use a different letter, say y, to represent the number of pears since they are two distinct measurements. When evaluating an expression, replace the variables with the input-values enclosed with ( ) s. Example A. a. Evaluate x if x = 6. When evaluating, insert the input enclosed in a ( ). Therefore, set x = ( 6) we ve x ( 6) = 6 b. Evaluate x if x = 6. x ( 6) = 18 c. Evaluate x if x = 6. x (6) = (6) = 7 d. Evaluate xyz if x =, y =, z = 1. xyz ( )( )( 1) = e. Evaluate x y if x =, y =. x y ( ) ( ) = + = f. Evaluate x y if x = and y =. Replace x by () and y by ( ) in the expression, we have *() ( ) = * 9 = 1 9 = g. Evaluate x + ( 8 y) if x = and y =. Replace x by, y by ( ) in the expression, () + ( 8 ( )) = 9 + ( 8 + ) = 9 + ( 6) = = 7 8

29 h. Evaluate (a b)(b c) if a =, b =, c =. (a b)(b c) (() ( ))(( ) ( )) = ( + )( + ) = ()() = 10 i. Evaluate (b a) if a =, b =. (( ) ( )) = ( 6 + 1) = (6) = 6 j. Evaluate b ac if a =, b =, and c =. ( ) ( )() = Variables and Evaluation Exercise. Evaluate. A. x with the input 1. x =. x =. x =. x = 1/ B. y x with the input. x =, y = 6. x =, y = 7. x = 1, y = 8. x = ½, y = 6 C. ( x) with the input 9. x = 10. x = 11. x = 1. x = 1/ D. x with the input 1. x = 1. x = 1. x = x = 1/ E. x with the input 17. x = 18. x = 19. x = 1 0. x = ½ F. x x 1 with the input 1. x =. x =. x = 1. x = ½ G. y + x with the input. x =, y = 6. x =, y = 7. x = 1, y = 8. x = 1, y = 1/ H. x x + x 1 with the input 9. x = 1 0. x = 1 1. x =. x = ½ I. b a with the input. a = 1, b =. a =, b =. a =, b = 8 6. a =, b = 1 J. b ac with the input 7. a =, b =, c = 8. a =, b =, c = 9. a = 1, b =, c = 0. a =, b =, c = 9

30 K. a b with the input c d 1. a = 1, b =, c =, d =. a =, b =, c = 1, d =. a =, b =, c =, d = 0. a = 1, b =, c =, d = 1 L. (a b)(b c) with the input (c d)(d a). a = 1, b =, c =, d = 6. a =, b =, c = 1, d = 7. a =, b =, c =, d = 0 8. a = 1, b =, c =, d = 1 M. b a c if 9. a =, b =, c =. 0. a =, b =, c = 0

31 Chapter Linear Equations and Inequalities 1 Expressions A mathematical expression is a calculation procedure written using numbers, variables, and operation symbols. Example A. + x the sum of and times x x x the difference between times the square of x and times x ( x) the square of the difference between and twice x The simplest type of expressions are linear expressions. These are expressions of the form ax + b where the a and b are numbers. Example B. x, 6 + x, are linear expressions, x 1, x are not linear. Combining Expressions Given an expression, each addend is called a term. There are two terms in ax + b. the x-term the number term or the constant term Just as apples + apples = apples we may combine two x-terms. Hence x + x = x. We combine positive x-terms and negative x-terms the same way we combine signed numbers. Hence x x = 8x. The number-terms may be combined such as 8 =. However just as apple + banana = apple + banana, (i.e. they can t be combined), x-terms may not be combined with number terms because they are different type of items. Hence + x stays as + x (not x). Example C. Combine. x + 9 x = x x + 9 = x + For the x-term ax, the number a is called the coefficient of the term. When we multiply a number with an x-term, we multiply it with the coefficient. Hence, (x) = (*)x =1x, and ( x) = ( )( )x = 8x. We may multiply a number with an expression and expand the result by the distributive law. Distributive Law A(B ±C) = AB ±AC = (B ±C)A Example D. Expand then simplify. a. (x ) = (x) ( )() = 10x + 0 1

32 b. (x ) + ( x) = 6x x = x Distributive law gives us the option of expanding the content of the parentheses i.e. extract items out of containers. Example E. A store sells two types of gift boxes Regular and Deluxe. The Regular has 1 apples and 8 bananas, the Deluxe has apples and bananas. We have boxes of Regular and boxes of Deluxe. How many apples and bananas are there? Let A stands for apple and B stands for banana, then Regular = (1A + 8B) and Deluxe = (A + B). Three boxes of Regular and four boxes of Deluxe is (1A + 8B) + (A + B) = 6A + B + 96A + 96B = 1A + 10B Hence we have 1 apples and 10 bananas. We usually start with the innermost set of parentheses to simplify an expression that has multiple layers of parentheses. In mathematics, ( ) s, [ ] s, and { } s are often used to distinguish the layers. This may not be the case for calculators or software where [ ] and { } may have other meanings. Always simplify the content of a set of parentheses first before expanding it. Example F. Expand. { x [ ( x 6)] } = { x [ + 8x + 1] } = { x [17 + 8x] } = { x 17 8x } expand, simplify, expand, simplify, = { 11x 1} expand, = x + 6 X-terms with fractional coefficients may be written in two ways, p q x or px p p ( but not as ) since q qx q x = p x q 1 = px * q Hence is the same as x x. Example G. Evaluate x if x = 6. Set x = (6) in we get x, (6) = 8 We may use the multiplier method to combine fraction terms i.e. multiply the problem by the LCD and divide by the LCD. Example H. Combine x + x x + x Multiply and divide by their LCD =1, = ( x + x)1 / 1 expand and cancel the denominators, = (* x + *x) / 1 = (16x + 1x) /1 = 1x 1

33 Expressions Exercise A. Combine like terms and simplify the expressions. 1. x + x. x x. x x. x + x. x + x + 6. x + x 7. 8 x 8. 8 x x 9. 8x x x x x 11. A + B A + B 1. 8B + A 9A B B. Expand then simplify the expressions. 1. (x + ) 1. (x ) 1. ( x ) 16. (6 + x) 17. (x + ) + (x ) 18. (x + ) ( x ) 19. 9(x 6) + ( + x) 0. 1(x + ) ( 7 x) 1. 7(8 6x) + ( x + ). ( 1x + ) ( 7x ). 7( 8 6x) ( x). ( 1x ) 6( 9x ). (A + B) (A + B) 6. 8(B + A) + 9(A B) 7. 11(A B) (A 1B) 8. 6(B 7A) 8(A B) C. Starting from the innermost ( ) expand and simplify. 9. x + [6 + ( + x)] 0. [ x ( 7 x)] [( x + ) + 6x] + x. 1x + [x ( x + 1)]. 7x + {8 [6(x ) ] x}. {8 [6(x ) ] x} [x ( x + )]. [( x) 6x] {x [x ( x + )]} D. Combine using the LCD-multiplication method 6. x + x 7. x x x 8 x x 6 x 0. Do 6 9 by the cross multiplication method. 1. As in example C with gift boxes Regular and Deluxe, the Regular contains1 apples and 8 bananas, the Deluxe has apples and bananas. Joe has 6 Regular boxes and 8 Deluxe boxes. How many of each type of fruit does he have?. As in example C with gift boxes Regular and Deluxe, the Regular contains1 apples and 8 bananas, the Deluxe has apples and bananas. For large orders we may ship them in crates or freight-containers where a crate contains 100 boxes Regular and 80 boxes Deluxe and a container holds 10 Regular boxes and 100 Deluxe boxes. King Kong ordered crates and containers, how many of each type of fruit does King Kong have?

34 Linear Equations I We left the house with a certain amount of money, say $x. After we bought a hamburger for $, we had $(x ) left. Suppose there was $1 remained, how much did we have in the beginning? In symbols, we've the equation x =1. Calculate backwards by adding to1, we get 1 + = 18. In symbols x =1 add to both sides + + so x = 18 We bought apples for $0. How much was each apple? Again, let x be the price of an apple, then apples cost $x. Since the total cost was 0, by dividing 0/ = and we get that each apple was $. In symbols, x = 0 both sides divided by so x 0 = so x = We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation. Example A. For 1 = x + 8, x = () is a solution because 1 = () + 8? 1 = 6 + 8? 1 = 1 yes Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x. These are the one-step equations. For example, x = 1, 1 = x +, *x = 1, 1 = x are one-step equations.

35 Basic principle for solving one- step-equations: To solve one-step-equations, isolate the x on one side by applying the opposite operation to both sides of the equation. Example B. Solve for x a. x = 1 Add to both sides + + x = 1 check: 1 =? 1 1 = 1 (yes) b. x + = 1 Subtract from both sides? x = 1 check: 1 + = 1 1 = 1 (yes) c. x = 1 Both sides divided by x = 1? x = check: () = 1 1 = 1 (yes) d. x = 1 Multiply both sides by ( x = 1) () x = 6 Check: 6 = 1 Fact: Given a linear equation if we +,, *, /, to both sides by the same quantity, the new equation will have the same solution. Next we tackle equations that require two steps. These are the ones that we have to collect the x-terms or the constant terms first. We do this by addition or subtraction. Example C. Solve for x a. x 6 = x Collect the x's by subtracting x from both sides x x 6 x Divide by = = x b. x 6 = collect the numbers by adding x = 9 div. by x/ = 9/ x = The more general linear equations have the form #x ±# = #x ±#, where # can be any number. We solve it by following steps: 1. Add or subtract to move the x-term to one side of the equation and get: #x ±# = # or # = #x ±#. Add or subtract the # to separate the number-term from the x-term to get: #x = # or # = #x.. Divide or multiply to get x: x = solution or solution = x The idea is to use + or, to separate the x-term from the number term, then use * or / to find the value of x.

36 Example D. Solve x = x + x x = x + 6 = x 6 x = = x subtract x to remove the x from one side. subtract to move the to the other side. divide by get x. Finally, all linear equations can be reduced to the format #x ±# = #x ±# by simplifying each side by itself first. Example E. Solve the equation. x (1 x) = (x 6) 1 Simplify each side by combining like-term first. x (1 x) = (x 6) 1 x + 9x = x x = x 19 x x 8x = x = 16 8x = 8 x = - gather the x s, subtract x from both sides gather the number-terms, + to both sides divide by 8 to get x. Linear Equations I Exercise A. Solve in one step by addition or subtraction. 1. x + =. x 1 =. = x. x + 8 = 1. x = 1/ 6. = x 1 B. Solve in one step by multiplication or division. 7. x = 8. x = 1 9. = x x = = x 1. 7 = x 1. x = 1. 7 = x 1. x = 7 C. Solve by collecting the x s to one side first. (Remember to keep the x s positive.) 16. x + = x 17. x 1 = x x = x x = x 0. x = 6 x 1. x + = + x. x 1= 6x. x + 7 = x. x + = 9 + x 6

37 D. Solve for x by first simplifying the equations to the form of #x ± # = #x ± #.. (x + ) = (x 1) 6. (x 1) + = x 9 7. (x ) = ( x 1) + x 8. (x + ) = (x 1) 9. x + (x ) = (x 1) 0. (x ) + = (x 1) + x (x + ) (x+ 1) = (x 1). x (x ) = (x 1) ( x). ( x) = (x + 1) 1. (x ) ( x) = (x +) + (x + 1). ( x) + ( x) = (x 1) + ( x) 6. 6(x ) (x +) = x + 6( x ) 8 In this section we make some remarks about solving equations. Recall that to solve linear equations we simplify each side of equation first then we separate the x's from the numbers. I. (Change-side-change-sign Rule) When gathering the x's to one side of the equation and the number terms to the other side, just move terms across to other side and switch their signs (instead of adding or subtracting to remove terms). x + a = b x = b a change-side change-sign Linear Equations II or x a = b x = b + a change-side change-sign There are two sides to group the x's, the right hand side or the left hand side. It s better to collect the x's to the side so that it ends up with positive x-term. 7

38 Example A. Solve. a. = x = x 11 = x b. (x + ) x = ( + x) + x x + 6 x = + x + x x + 6 = + x simplify each side move x and to the other + 6 = x x sides, change their signs 10 = x 1 move to the other side, 1 change its sign x = x = 1 x = c. + x = move +6 to the left, it turns into 6 Rule II. If x = c, then x = c Example B. a. If x = 1 then x = 1 b. If x = then x = Recall that an equation or a relation expressed using fractions may always be restated in a way without using fractions. Example C. The value of / of an apple is the same as the value of ¾ of an orange, restate this relation in whole numbers. We've A = G, LCD =1, multiply 1 to both sides, ( A = G ) *1 8A = 9G or the value of 8 apples equal to 9 oranges. III. (Fractional Equations Rule) Multiply fractional equations by the LCD to remove the fractions first, then solve. Example D. Solve the equations a. x = 6 clear the denominator, multiply both sides by ( x = 6) x = div by x = 8 1 x b. + = 6 x the LCD = 1, multiply it to both sides to clear the denominators ( 1 x + = 6 x ) *1 x + 8 = 10x 9 move x to the right and 9 to the left and switch signs = 10x x 17 = 7x div by 7 17 = x 7 8

39 c. 1 (x 0) = x 7 multiply 100 to both sides to remove denominators 1 [ 1 1 (x 0) = x ] * (x 0) = x 700 1x 00 = x = x 1x 00 = 0x 00 / 0 = x 80 = x d. 0.(x 100) = 0.10x 1 Change the decimal into fractions, we get (x 100) = 100 x 1 multiply 100 to both sides to remove denominators You finish it Ans: x =160 IV. (Reduction Rule) Use division to reduce equations to simpler ones. Specifically divide the common factor of the coefficients of each term to make them smaller and easier to work with. Example. E. Simplify the equation first then solve. 1x 9 = 70x 98 divide each term by 7 1x 9 70x 98 = x 7 = 10x = 10x x 7 = 8x 7 8 = x We should always reduce the equation first if it is possible. 9

40 There re two type of equations that give unusual results. The first type is referred to as identities. A simple identity is the equation x = x. This corresponds to the trick question What number x is equal to itself? The answer of course is that x can be any number or that the solutions of equation x = x are all numbers. This is also the case for the any equation where both sides are identical such as x + 1 = x + 1,1 x = 1 x etc An equation with identical expressions on both sides or can be rearranged into identical sides has all numbers as its solutions. Such an equation is called an identity. Example F. Solve. (x 1) + = x ( x 1) expand x + = x + x + 1 simplify x + 1 = x + 1 two sides are identical So this equation is an identity and every number is a solution. At the opposite end of the identities are the impossible equations where there is no solution at all. An example is the equation x = x + 1. This corresponds to the trick question What number is still the same after we add 1 to it? Of course there no such number. If we attempt to solve x = x + 1 x x = 1 we get 0 = 1 which is an impossibility. These equations are called inconsistent equations. Example F. Solve the equation (x 1) + = x ( x 1) expand x + = x + x + 1 simplify x + = x + 1 = 1 So this is an inconsistent equation and there is no solution. Linear Equations II Exercise. A. Solve for x using the switch-side-switch-sign rule. Remember to move the x s first and get positive x s. 1. x + = x. x 1 = x 7. x = x 8 x = x = 7 = x. x = x. x = 6 x 6. x + = + x 7. x 1= 6x 8. x + 7 = x 9. x + = 9 + x B. Solve the following fractional equations by using the LCD to remove the denominators first. x 10. x = 11. x = = x = 1. x = x = 6 0

41 x + = x 17. x 1 6 = 8 x x = 7 10 x x = 16 x (x 0) = 0 x (x + ) = 0 (x ) (x +1) = x (0 x) + = 0 (x 0) x 0. = 1 0.6x. 0.1x 0.1x = 0.x x = 0.19x x 0.11 = 0. 0.x C. Reduce the equations then solve. 8. x 1 = 0 6x 9. 1x 10x = x 0 0. (x ) = 1(x + ) 8x 1. 1x 10(x + ) = x 0 D. Identify which equations are identities and which are inconsistent.. x + 1 = x +. x + = ( x). x + 1 = x (x + 1). (x + ) = x (7x ) 6. (x ) = 1 (1 x) 7. (x ) = 1 (1 x) Linear Word-Problems The following are some key words translated into mathematical operations. +: add, sum, plus, total, combine, increased by, # more than.. : subtract, difference, minus, decreased by, # less than.. *: multiply, product, times, fractions or % of the amount.. /: divide, quotient, shared equally, ratio.. Twice = Double = * (amount) Square = (amount) Example A. Let x be an unknown number, write the mathematical expressions represent the following amounts. a.00 more than twice x x + 00 b. / of x x c. 0% of the sum of x and ( x+1000) 100 1

42 Follow the following steps to solve word problems. Identify the unknown quantities. Set x to be the unknown quantity where all the other unknown quantities in the problem are described in terms of it. Express them in terms of the chosen x. Organize information into a table if possible. Translate the problem into an equation. Solve the equation. Example A. Mary and Joe share $10, Mary gets $9 more than Joe. How much does each get? Let x = $ Joe gets, then Mary gets (x + 9). They ve $10 in total. x + (x + 9) = 10 x + 9 = 10 x = 10 9 x = 1 x = ½ (for Joe) 9 + ½ = 9½ (for Mary) Example B. Mary and Joe are to share $10, Joe gets / of what Mary gets. How much does each get? Mary gets $ x, so Joe gets $ x Together they have $10, so x + x = 10 multiply both sides by ( x + x = 10) * x + x = 60 x = 60 x = 60/ x = 7 Therefore Mary gets $7 and the rest 10 7 = $8 is the amount for Joe. Concentration Formula The concentration of a chemical in a solution is usually given in %. Let r = concentration (in %) S = amount of solution C = amount of chemical Then r*s = C Example C. We have 60 gallons brine (salt water) of % concentration. How much salt is there? r = concentration = % = 100 S = amount of brine = 60 gal C = amount of salt =? * rs = 100 * 60 = = 1 gallons of salt

43 Example D. (Mixture problem) How many gallons of 10% brine must be added to 0 gallons of 0% brine to dilute the mixture to 0% brine? Make the standard table for mixture problems concentration # of gallons salt (r*s) 10 10% x 100 x 0 0% * 0 0 0% (x+0) (x+0) Total amount of salt in the final 100 mixture Hence x * 0 = (x + 0) [ x * 0 = (x + 0) ] * 100 multiply by 100 to clear denominator. 10x = 0(x+0) 10x = 0x = 0x 10x 600 = 10x 60 = x Ans: 60 gallons Simple Interest Formula The interest rate of an investment is usually given in %. Let r = interest rate (in %) P = principal I = amount of interest for one year then r*p = I. Example E. We have $000 in a saving account that gives 6% annual interest. How much interest will be there after one year? r = interest rate = P = principal I = amount of interest Hence r*p = * 000 = 6(0) = $180 = interest. Remark: The general formula for simple interest is P*r*t = I where t is the number of years. Note the similarity to the concentration formula. Example F. (Mixed investments) We have $000 more in a 6% account than in a % account. In one year, their combined interest is $60. How much is in each account? We make a table. Let x = amount in the % account. rate principal interest 6 6% (x + 000) 100 (x+000) Total amount % x } 100 x of interest is $60 6 Hence: 100 (x + 000) x = 60 [ 6 1 ] * (x + 000) x = (x + 000) + x = x x = x = = 000 x = 000/11 = $000 So we've $000 in the % accnt. and $6000 in the 6% accnt.

44 Linear Word-Problems Exercise A. Express each expression in terms of the given x. Let x = $ that Joe has. 1. Mary has $10 less than what Joe has, how much does Mary have?. Mary has $10 more than twice what Joe has, how much does Mary have?. Mary has twice the amount that s $10 more than what Joe has, how much does Mary have?. Together Mary and Joe have $100, how much does Mary have?. Mary has $100 less than what Joe has, what is twice what Joe and Mary have together? Simplify your answer. 6. Mary has $100 more than Joe, what is half of what Joe and Mary have together? Simplify your answer. B. Peanuts cost $/lb, cashews cost $8/lb. We have x lbs of peanuts. Translate each quantity described below in terms of x. 7. What is the cost for the peanuts? 8. We have more lbs of cashews as peanuts, what is the cost for the cashews? 9. Refer to problem 7 and 8, what is the total cost for all the nuts? Simplify answer. 10. We have twice as many of cashews as peanuts in weight, what is the total cost of the nuts? Simplify your answer. 11. We have a total of 0 lb of peanuts and cashews, how many lbs of cashews do we have? What is the total cost for all the nuts? Simplify you answers. B. Money Sharing Problems. Select the x. Set up an equation then solve. 1. A and B are to share $10. A gets $0 more than B, how much does each get? 1. A and B are to share $10. A gets $16 less than three times of what B gets, how much does each get? 1. A, B and C are to share $10. A gets $0 more than what B gets, C gets $10 less than what B gets, how much does each get? 1. A, B and C are to share $10. A gets $0 more than what twice of B gets, C gets $10 more than what A gets, how much does each get? 16. A and B are to share $10. A gets / of what B gets, how much does each get? 17. A, B and C are to share $10. A gets 1/ of what B gets, C gets ¾ of what B gets, how much does each get? C. Wet Mixtures Problems (Make a table for each problem 8 19) 18. How many gallons of 0% brine must be added to 0 gallons of 0% brine to dilute the mixture to a % brine? 19. How many gallons of pure water must be added to 0 gallons of 0% brine to dilute the mixture to a 0% brine? 0. How many gallons of 10% brine must be added with 0% brine to make 0 gallons of 0% brine? 1. How many gallons of 0% brine must be added with pure water to make 0 gallons of 0% brine? D. Mixed Investment Problems.. We have $000 more saved in an account that gives % interest rate than in an account that gives % interest rate. In one year, their combined interest is $600. How much is in each account?. We have saved at a 6% account $000 less than twice than in a % account. In one year, their combined interest is $1680. How much is in each account?. The combined saving in two accounts is $1,000. One account gives % interest rate, the other gives % interest rate. The combined interest is %70 in one year. How much is in each account?. The combined saving in two accounts is $1,000. One account gives 6% interest rate, the other gives % interest rate. The combined interest is $70 in one year. How much is in each account? E. Dry Mixture Problem These are similar to the wet mixture problems. 6. Peanuts cost $/lb, cashews cost $6/lb. How many lbs of peanuts are needed to mix with1 lbs cashews to make a peanut cashews mixture that costs $/lb? 7. Peanuts cost $/lb, cashews cost $6/lb. How many lbs of each are needed to make 1 lbs of peanut cashews mixture that costs $/lb? 8. Peanuts cost $/lb, cashews cost $8/lb. How many lbs of cashews are needed to mix with 1 lbs of peanuts to get a peanut cashews mixture that costs $6/lb? 9. Peanuts cost $/lb, cashews cost $8/lb. How many lbs of each are needed to make 0 lbs of peanut cashews mixture that cost $/lb?

45 Literal Equations Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We do this, just as solving equations in x, by +,, *, / the same quantities to both sides of the equations. These quantities may be numbers or variables. Example A. a. Solve for x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c b b x = c b b. Solve for w if yw =. Remove y from the LHS by dividing both sides by y. yw = yw/y = /y w = y Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign. To solve for a specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions.. Isolate the term containing the variable we wanted to solve for move all the other terms to other side of the equation.. Isolate the specific variable by dividing the rest of the factor to the other side. Example B. a. Solve for x if (a + b)x = c (a + b) x = c div the RHS by (a + b) x = c (a + b) b. Solve for w if y w = t y w = t div the RHS by y w = t y c. Solve for a if b ac = b ac = ca = b a = b c move b to the RHS div the RHS by c d. Solve for y if a(x y) = 10 a(x y) = 10 expand ax ay = 10 subtract ax ay = 10 ax div by a y = 10 ax a

46 Multiply by the LCD to get rid of the denominator then solve. Example C. Solve for d if = = d + b d d = d + b ( ) d d d = d + b b = d + d b = 7d b = d 7 d + b d multiply by the LCD d move d and b div. by 7 Exercise. Solve for the indicated variables. 1. a b = d e for b.. a b = d e for e.. *b + d = e for b.. a*b + d = e for b.. ( + a)*b + d = e for b. 6. L + W = P for W 7. (x + 6)y = for y 8. x + 6y = for y 9. x + 6xy = for y 10. x (x + 6)y = z for y 11. w = t 6 for t 1. w = 6 t for t 1. w = t 6 for t 1. w = t b a for t 1. w = t a b for t 16. (x + 6)y = for x 17. w = t 6 + a for t 18. w = at b for t 19. w = at c b + d for t 0. = t t b for t 6 Inequalities We associate each real number with a position on a line, positive numbers to the right and negative numbers to the left π.1.. / ½ π.1.. This line with each position addressed by a real number is called the real (number) line. Given two numbers corresponding to two points on the real line, we define the number to the right to be greater than the number to the left. L R < + We write this as L < R and called this the natural form because it corresponds to their respective positions on the real line. This relation may also be written as R > L (less preferable). 6

47 Example A. <, <, 0 > 1 are true statements and <, < are false statements. If we want all the numbers greater than, we may denote them as "all number x where < x". In general, we write "a < x" for all the numbers x greater than a (excluding a). In picture, a a < x + open dot If we want all the numbers x greater than or equal to a (including a), we write it as a < x. In picture a a < x + solid dot The numbers x fit the description a < x < b where a < b are all the numbers x between a and b. A line segment as such is called an interval. a a < x < b b + Example B. a. Draw 1 < x <. It s in the natural form. Mark the numbers and x on the line in order accordingly. x b. Draw 0 > x > Put it in the natural form < x < 0. Then mark the numbers and x in order accordingly. x - 0 Expressions such as < x > or < x < do not have any solution. Adding or subtracting the same quantity to both retains the inequality sign, i.e. if a < b, then a ± c < b ± c. For example 6 < 1, then 6 + < 1 +. We use the this fact to solve inequalities. + Example C. Solve x < 1 and draw the solution. x < 1 add to both sides x + < 1 + x < 1 x 0 1 A number c is positive means that 0 < c. We may multiply or divide a positive number to the inequality and keep the same inequality sign, i.e. if 0 < c and a < b, then ac < bc. For example 6 < 1 is true, then multiplying by *6 < *1 or 18 < 6 is also true. Example D. Solve x > 1 and draw the solution. x > 1 divide by and keep the inequality sign x/ > 1/ x > or < x x

48 A number c is negative means c < 0. Multiplying or dividing by an negative number reverses the inequality sign, i.e. if c < 0 and a < b then ca > cb. For example 6 < 1 is true. If we multiply 1 to both sides then ( 1)6 > ( 1)1 6 > 1 which is true. Multiplying by 1 switches the left-right positions of the originals. 1 < < 1 Example E. Solve x + < and draw the solution. x + < subtract from both sides x < multiply by 1 to get x, reverse the inequality ( x) > x > or < x To solve inequalities: 1. Simplify both sides of the inequalities. Gather the x-terms to one side and the number-terms to the other sides (use the change side-change sign rule).. Multiply or divide to get x. If we multiply or divide by negative numbers to both sides, the inequality sign must be turned around. This rule can be avoided by keeping the x-term positive. Example F. Solve x + > x + 9 x + > x + 9 move the x and, change side-change sign x x > 9 x > div. x > x > or < x + 0 Example G. Solve ( x) > (x + 9) x ( x) > (x + 9) x simplify each side 6 x > x + 18 x 6 x > 18 move 18 and x (change sign) 6 18 > x 1 > x div. by (no need to switch >) 1 > x > x or x < - We also have inequalities in the form of intervals. We solve them by +,, *, / to all three parts of the inequalities. Again, we + or remove the number term in the middle first, then divide or multiply to get x. The answer is an interval of numbers

49 Example H. (Interval Inequality) Solve 1 > x + 6 > and draw. 1 > x + 6 > subtract > x > 10 div. by -, switch inequality sign 6 -x -10 < < < x < Inequalities Exercise. A. Draw the following Inequalities. Indicate clearly whether the end points are included or not. 1. x <. x. x < 8. x 1 B. Write in the natural form then draw them.. x 6. > x 7. x 8 8. x > 1 C. Draw the following intervals, state so if it is impossible > x 10. < x > x 8 1. < x 1. 6 > x 8 1. > x 1. 7 x x 9 D. Solve the following Inequalities and draw the solution. 17. x + < 18. x x + 1 < 8 0. x + 1 x 1. x + 1 x. (x + ) (x 1). (x 1) + x 9. (x ) > ( x 1) + x. (x + ) (x 1) 6. x + (x ) < (x 1) 7. (x ) + (x 1) + x + 1 E. Clear the denominator first then solve and draw the solution. 8. x 9. 7 > x 0. x < 1 x + 1. x. x > x 1. x. x + 7 > x. + 1 x 6. x < x 7. x x 6 F. Solve the following interval inequalities x < > x > 0. < x + < 1. 1 x x < > 1 x > 7 9

50 Chapter Lines and Linear Equations in x & y 1 The Rectangular Coordinate System A coordinate system is a system of assigning addresses for positions in the plane ( D) or in space ( D). The rectangular coordinate system for the plane consists of a rectangular grid where each point in the plane is addressed by an ordered pair of numbers (x, y). The horizontal axis is called the x-axis. The vertical axis is called the y-axis. The point where the axes meet is called the origin. Starting from the origin, each point is addressed by its ordered pair (x, y) where: x = amount to move right (+) or left ( ). y = amount to move up (+) or down ( ). R B A C P Q x = amount to move right (+) or left ( ). y = amount to move up (+) or down ( ). For example, the point P corresponds to (, ) is right, and down from the origin. Example A. Label the points A(-1, ), B(-, -),C(0, -). The ordered pair (x, y) corresponds to a point is called the coordinate of the point, x is the x-coordinate and y is the y-coordinate. Example B: Find the coordinate of P, Q, R as shown. P(, ), Q(, -), R(-6, 0) Q Q1 (,+) (+,+) Q Q (, ) (+, ) The coordinate of the origin is (0, 0). Any point on the x-axis has coordinate of the form (x, 0). Any point on the y-axis has coordinate of the form (0, y). The axes divide the plane into four parts. Counter clockwise, they are denoted as quadrants I, II, III, and IV. Respectively, the signs of the coordinates of each quadrant are shown. 0

51 (,) (, ) (,) (, ) When the x-coordinate of the a point (x, y) is changed to its opposite as ( x, y), the new point is the reflection across the y-axis. When the y-coordinate of the a point (x, y) is changed to its opposite as (x, y), the new point is the reflection across the x-axis. ( x, y) is the reflection of (x, y) across the origin. x coord. decreased by C (, ) x coord. increased by A B (, ) (6, ) Movements and Coordinates Let A be the point (, ). Suppose it s x coordinate is increased by to ( +, ) = (6, ) - to the point B, this corresponds to moving A to the right by. Similarly if the x coordinate of (, ) is decreased by to (, ) = (, ) - to the point C, this corresponds to moving A to the left by. Hence we conclude that changes in the x coordinates of a point move the point right and left. If the x change is +, the point moves to the right. If the x change is, the point moves to the left. y coord. increased by y coord. decreased by D (, 7) A (, ) E (, 1) Again let A be the point (, ). Suppose its y coordinate is increased by to (, + ) = (, 7) - to the point D, this corresponds to moving A up by. Similarly if the y coordinate of (, ) is decreased by to (, ) = (, 1) - to the point E, this corresponds to moving A down by. Hence we conclude that changes in the y coordinates of a point move the point right and left. If the y change is +, the point moves up. If the y change is, the point moves down. 1

52 Example. C. a. Let A be the point (, ). What is the coordinate of the point B that is 100 units directly left of A? Moving left corresponds to decreasing the x-coordinate. Hence B is ( 100, ) = ( 10, ). b. What is the coordinate of the point C that is 100 units directly above A? Moving up corresponds to increasing the y-coordinate. Hence C is (, ) = (, +100) = (, 10). c. What is the coordinate of the point D that is 0 to the right and 0 below A? We need to add 0 to the x coordinate (to the right) and subtract 0 from the y coordinate (to go down). Hence D has coordinate ( + 0, 0) = (8, ). The Rectangular Coordinate System Exercise. A. a. Write down the coordinates of the following points. F C E B A G D H B. Plot the following points on the graph paper.. a. (, 0) b. (, 0) c. (, 0) d. ( 8, 0) e. ( 10, 0) All these points are on which axis?. a. (0, ) b. (0, ) c. (0, ) d. (0, 6) e. (0, 7) All these points are on which axis?. a. (, ) b. (, ) c. (1, 7) d. (7, 1) e. (6, 6) All these points are in which quadrant?. a. (, ) b. (, ) c. ( 1, 7) d. ( 7, 1) e. ( 6, 6) All these points are in which quadrant? 6. List three coordinates whose locations are in the nd quadrant and plot them. 7. List three coordinates whose locations are in the th quadrant and plot them.

53 Linear Equations and Lines In the rectangular coordinate system, ordered pairs (x, y) s correspond to locations of points. Collections of points may be specified by the mathematics relations between the x-coordinate and the y coordinate. The plot of points that fit a given relation is called the graph of that relation. To make a graph of a given mathematics relation, make a table of points that fit the description and plot them. Example A. Graph the points (x, y) where x = (y can be anything). Make a table of ordered pairs of points that fit the description x =. x y 0 6 Graph of x = Example B. Graph the points (x, y) where y = x. Make a table of points that fit the description y = x. To find one such point, we set one of the coordinates to be a number, any number, than use the relation to find the other coordinate. Repeat this a few times. x y Graph the points (x, y) where y = x First degree equation in the variables x and y are equations that may be put into the form Ax + By = C where A, B, C are numbers. First degree equations are the same as linear equations. They are called linear because their graphs are straight lines. To graph a linear equation, find a few ordered pairs that fit the equation. To find one such ordered pair, assign a value to x, plug it into the equation and solve for the y (or assign a value to y and solve for the x). For graphing lines, find at least two ordered pairs. Example C. Graph the following linear equations. a. y = x Make a table by selecting a few numbers for x. For easy caluation we set x = -1, 0, 1, and. Plug each of these value into x and find its corresponding y to form an ordered pair.

54 For y = x : If x = -1, then y = (-1) = -7 If x = 0, then y = (0) = - If x = 1, then y = (1) = - If x =, then y = () = -1 x y b. -y = 1 Simplify as y = - Make a table by selecting a few numbers for x. However, y = - is always. c. x = 1 Simplify as x = 6 Make a table. However the only selction for x is x = 6 and y could be any number. x y x y Summary of the graphs of linear equations: a. y = x b. -y = 1 c. x = 1 If both variables x and y are present in the equation, the graph is a tilted line. If the equation has only y (no x), the graph is a horizontal line. If the equation has only x (no y), the graph is a vertical line.

55 The x-intercepts is where the line crosses the x-axis. We set y = 0 in the equation to find the x-intercept. The y-intercepts is where the line crosses the y-axis. We set x = 0 in the equation to find the y-intercept. Since two points determine a line, an easy method to graph linear equations is the intercept method, i.e. plot the x-intercept and the y intercept and the graph is the line that passes through them. Example C. Graph x y = 1 by the intercept method. If x = 0, we get x y (0) y = y-int so y = - If y = 0, we get 6 0 x-int x (0) = 1 so x = 6 Linear Equations and Lines Exercise. A. Solve the indicated variable for each equation with the given assigned value. 1. x + y = and x = 1, find y.. x y = and y = 1, find x.. x = 6 and y = 1, find x.. y = and x =, find y.. y = x and x =, find y. 6. y = x + and x =, find y. 7. x y = 1 and y =, find x. 8. x = 6 y and y =, find x. 9. y = x and x =, find y. 10. x + y = and x = 0, find y. 11. x + y = and y = 0, find x. 1. x y = 1 and x = 0, find y. 1. x y = 1 and y = 0, find x = x y and y =, find x. B. a. Complete the tables for each equation with given values. b. Plot the points from the table. c. Graph the line. 1. x + y = 16. y = x = y = x x y x y x y x y x y = 0. y = 6 + x x y x y y = 6 x y 0 1. y + x =1 x y 0 0 1

56 C. Make a table for each equation with at least ordered pairs. (remember that you get to select one entry in each row as shown in the tables above) then graph the line.. x y =. x = 6. y 7= = 8 x 7. y = x + 8. x = 6 9. x = 6 y 0. y 1 = x 1. x + y =. 6 = x y. x = y. x + y = 10. For problems 9, 0, 1 and, use the intercept-tables as shown to graph the lines. 6. Why can t we use the above intercept method to graph the lines for problems, 6 or? 7. By inspection identify which equations give horizontal lines, which give vertical lines and which give tilted lines. x 0 0 y intercept-table In general, we need one piece of information to solve for one unknown quantity. If hamburgers cost dollars and x is the cost of a hamburger, then x = or x = $. If there are two unknowns, we need two pieces of information to solve them, three unknowns needs three pieces of information, etc These lead to systems of equations. A system of linear equations is a collection two or more linear equations in two or more variables. A solution for the system is a collection of numbers, one for each variable, that works for all equations in the system. For example, x + y = 7 x + y = is a system. x = and y = is a solution because the ordered pair fits both equations. This solution is written as (, ). { Systems of Linear Equations Example A. Suppose two hamburgers and a salad cost $7, and one hamburger and one salad cost $. Let x = cost of a hamburger, y = cost of a salad. We can translate these into the system. x + y = 7 E1 { x + y = E The difference between the order is one hamburger and the difference between the cost is $, so the hamburger is $. In symbols, subtract the equations, i.e. E1 E: x + y = 7 ) x + y = x + 0 = so x = substitute for x back into E, we get: + y = or y = Hence the solution is (, ) or, $ for a hamburger, $ for a salad. 6

57 Example B. Suppose four hamburgers and three salads is $18, three hamburger and two salads is $1. How much does each cost? Let x = cost of a hamburger, y = cost of a salad. We translate these into the system: x + y = 18 E1 { x + y = 1 E Subtracting the equations now will not eliminate the x nor y. We have to adjust the equations before we add or subtract them to eliminate one of the variable. Suppose we chose to eliminate the y-terms, we find the LCM of y and y first. It's 6y. Then we multiply E1 by, E by (-), and add the results, the y-terms are eliminated. E1*: 8x + 6y = 6 E*(-): + ) -9x 6y = -9 -x = - x = To find y, set for x in E: x + y = 1 and get () + y = y = 1 y = y = Hence the solution is (, ). Therefore a hamburger cost $ and a salad cost $. The above method is called the elimination (addition) method. We summarize the steps below. 1. Line up the x s & y s to the left and the numbers to the right.. Select the variable x or y to eliminate.. Find the LCM of the terms with that variable, and convert these terms to the LCM (by multiplication).. Add or subtract the adjusted equations and solve the resulting equation.. Substitute the answer back into any equation to solve for the other variable. { x + y = Example C. Solve E1 x y = -1 E Eliminate the y. The LCM of the y-terms is 1y. Multiply E1 by and E by. *E1: 1x + 1y = 6 *E: ) 8x 1y = - Add: x + 0 = - 6 x = - Set - for x in E1 and get (-) + y = y = y = 1 y = Hence the solution is (-, ). In all the above examples, we obtain exactly one solution in each case. However, there are two other possibilities. The system might not have any solution or the system might have infinite many solutions. 7

58 Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = (E1) x + y = (E) Remove the x-terms by subtracting the equations. E1 E : x + y = ) x + y = 0 = -1 This is impossible! Such systems are said to be contradictory or inconsistent. These system don t have solution. Two Special Cases: I. Contradictory (Inconsistent) Systems Example D. { x + y = (E1) x + y = (E) Remove the x-terms by subtracting the equations. E1 E : x + y = ) x + y = 0 = -1 This is impossible! Such systems are said to be contradictory or inconsistent. These system don t have solution. Systems of Linear Equations Exercise. A. Select the solution of the system. 1. a. (0, ) b. (1, ) c. (, 1) d.(, 0). a. (0, ) b. (1, ) c. (, 1) d.(, 0). a. (0, ) b. (1, ) c. (, 1) d.(, 0). a. (0, ) b. (1, ) c. (, 1) d.(, 0) B. Solve the following systems.. { y = x + y = 6. { x = 1 x + y = 7. { y = 1 x + y = { x + y = x + y = { x + y = x + y = { x + y = x + y = { x + y = x + y = 6 C. Add or subtract vertically. (These are warm ups) 8. { x = x + y = x y y 1. +x +) 9x ) 9 ) 9y ) 9 +) y ) x 8

59 D. Solve the following system by the elimination method. 16. { x + y = 17. { x + y = x y = 6 x y = x + y = { 0. x y = 6 { x + y = x 9y =. x + y = 1 {. x y = x + y = { x y = 1. { x + y = x + y = 18. x + y = 1 { x + y = 1. x + y = 1 { x + y =. x + y = { x + y = x y = { 1 x 1 y = 1. x + y = 10 { x + y = 7 6. x + 9y = 1 { x y = 8 1 x + 1 y = { x 1 y = 1 6 E. Which system is inconsistent and which is dependent? 9. { x + y = x + 6y = 8 0. { x y = 8x y = 6 Systems of Linear Equations II There are two other methods to solve system of equations. Substitution Method In substitution method, we solve for one of the variables in terms of the other, then substitute the result into the other equation. Example A. Solve x + y = 7 E1 { x + y = E by the substitution method. From E, x + y =, we get x = y. Then we replace x by ( y) in E1 and get ( y) + y = 7 10 y + y = 7 10 y = 7 = y Put = y back to E to find x, we get x + = x = Therefore the solution is (, ) We use the substitution method when it's easy to solve for one of the variable in terms of the other. Specifically, it is easy to solve for a variable when an equation contains an single x or single y. x y = 7 E1 Example B. Solve { x + y = 7 E by the substitution method. By inspection, we see that it's easy to solve for the y using E1. From E1, x y = 7, we get x 7= y. Substitute y by (x 7) in E and get x + (x 7) = 7 x + x 1 = 7 7x = 1 x = To find y, use the substitution equation, set x = in y = x 7 y = () 7 y = -1 Hence the solution is (, -1). 9

60 Graphing Method Given a system of linear equations the graph of each equation is a straight line. The solution of the system is the intersection point (x, y) of the these lines. Thus we may find the solution by graphing the lines and locate the point of intersection graphically. In general, we don't not use the graphing method because it is not easy to do it accurately. x + y = 7 E1 Example C. Solve { graphically. x + y = E Use intercept method. x y 0 7 7/ 0 x + y = 7 x + y = (Check another point.) x y 0 0 (0, 7) (0, ) (7/, 0) The intersection (, ) is the solution E1 (, 0) E E1 Graphing of inconsistent systems are parallel lines which do not intersect, hence there is no solution. x + y = 7 E1 Example D. Solve { graphically. x + y = E x + y = 7 x y x + y = x y 0 0 (0, ) (0, 7) (, 0) No intersection. No solution (7, 0) Graphs of dependent systems are identical lines, hence every point on the line is a solution. x + y = E1 Example E. Solve { graphically. x + y = 10 E x + y = 10 is the same as x + y = x y 0 0 (0, ) (, ) Every point is a solution, e.g.(0, ), (, ), (, 0) (, 0) 60

61 1. { y = x x + y =. x + y = 1 { y = x Systems of Linear Equations II Exercise. Solve by the substitution method.. { x + y =. { x = y x + 6 = y x y = 6. x + y = { x = 6 + y 10. Graph the inconsistent system { x y = 8x y = Graph the dependent system { x + y = x + 6y = 8 6. { x = y x 9y = Problem 7, 8, 9: Solve problem 1,, and by graphing. Linear Word-Problems II Following are key-words translated into mathematic operations. +: add, sum, plus, total, combine, increased by, # more than.. : subtract, difference, minus, decreased by, # less than.. *: multiply, product, times, fractions or % of the amount.. /: divide, quotient, shared equally, ratio.. Twice = Double = * (amount) Square = (amount) In chapter, we solved problems with a single variable. Many of those problems may be solved using two variables and a system of equations. The advantage of using two variables instead of one is that it's easier to construct the equations to form the system. To do this: I. identify the unknown quantities and named them as x and y II. usually there are two numerical relations given, translate each numerical relation into an equation to make a system III. solve the system. Example A. A 0-foot rope is cut into two pieces. The longer one is 6 feet less than twice of the short piece. How long is each piece? Let x = length of the long piece. y = length of the short piece Then x + y = E1 { x = y E Use the substitution method. Substitute x = (y 6) into E1, we get (y 6) + y = 0 solve for y y 6 + y = 0 y 6 = 0 y = 6 y = 6/ = 1 Put y = 1 into E, we get x = (1) 6 =18. Hence the two pieces are 18ft and 1ft. 61

62 Example B. Suppose four hamburgers and three fries costs $18, and three hamburgers and two fries costs $1. Find the price of each. Let x = price of a hamburger, y = price of fries The system is x + y = 18 E1 { x + y = 1 E Use the elimination method. LCM of y and y is 6y. Multiply E1 by and E by, subtract *E1: 8x + 6y = 6 *E: ) 9x + 6y = 9 x = x = To find y, put x = into E, () + y = y = 1 y = 1 9 y = y = / =. So a hamburger is $, and fries are $. Following problems use formulas in setting up the equations. The RTD Formula Let R = Rate or speed (usually mph) T = Time (in hours) D = Distance (in miles) then RT = D. For example, if we drove from LA to SF at a rate of 7 mph and it took 6 hours, it means R = 7, T = 6. Therefore 7(6) = 0 miles is the distance between the two cities. Up-and-down-stream-problems When a boat travels up and down a stream, the speed of the boat depends on the boat's speed R in still water (cruising speed) and the current's speed C. Going upstream, the current slows the boat down and going downstream it speeds up the boat. In fact, the upstream speed is (R C), and the downstream speed is (R + C). For example, a boat that travels mph in still water goes upstream against a 1 mph current, it's upstream rate is mph. It would take 6 hours to go 1 miles. If it travels downstream, it can travel at mph and it would take hours to go 1 miles. Example C. A boat can travel upstream for miles in 6 hours and the same distance downstream in hours. What is the cruising speed of the boat and the current speed? Let R = cruising speed, C = current speed, Make a table. Rate Time (T) Distance (D) Upstream R C 6 Downstream R + C By the formula R*T = D we get two equations. 6

63 { 6(R C) = --- E1 (R + C) = --- E Simplify each equation, divide E1 by 6 and divide E by. { R C = --- E R + C = E Add the equations to eliminate C +) R = 1 R = 6 Substitute R = 6 into E, 6 + C = 8 C = So, R = cruising speed = 6 mph, C = current speed = mph Simple Interest Formula Recall the simple interest of an investment is I = R*P where I = amount of interest for one year R = interest (in %) P = principal Example D. (Mixed investments) We have $0,000 saved in two accounts which give % and 6% interest respectively. In one year, their combined interest is $110. How much is in each account? Make a table using the interest formula R*P = I rate principal interest 6% x % y * x 100 y * total 0, { x + y = 0, E1 x + y = E 100 Multiply E by 100 to clear the denominator. 6 ( 100 x y = 110) x + y = E { x + y = 0, E1 6x + y = E To eliminate y, multiply E1 by - and add to E. E1(-): -x y = -100,000 6x + y = ) x = 1000 Sub x=1000 into E1, y = 0000 y = 000 Linear Word-Problems II So there are $1,000 in the 6% account, and $,000 in the % account. 6

64 Linear Word-Problems II Exercise. A. The problems below are from the word problems for linear equations with one variable. 1. A and B are to share $10. A gets $0 more than B, how much does each get? (Just let A = $ that Mr. A has, and B = $ that Mr. B has.). A and B are to share $10. A gets $0 more than twice of what B gets, how much does each get?. A and B are to share $10. A gets $16 less than three times of what B gets, how much does each get? Let x = number of lb of peanuts y = number of lb of cashews Make tables for problems Peanuts costs $/lb, cashews costs $8/lb. How many lbs of each are needed to get 0 lbs of peanut cashews mixture that s $/lb?. Peanuts costs $/lb, cashews costs $6/lb. How many lbs of each are needed to get 1 lbs of peanut cashews mixture that s $/lb? 6. We have $000 more saved at an account that gives % interest than at an account that gives % interest. In one year, their combined interest is $600. How much is in each account? 7. We have saved at a 6% account $000 less than twice than at a % account. In one year, their combined interest is $0. How much is in each account? 8. The combined saving in two accounts is $1,000. One account gives % interest, the other gives % interest. The combined interest is %70 in one year. How much is in each account? B. Find x = $ of cashews and y = $ of peanuts per lb, solve for x and y with the following information. 9. Order 1. (in lb) Order (in lb) 10. Cashews Peanuts 1 Cost $16 $17 Order 1. (in lb) Cashews Peanuts 1 Cost $17 $1 Order (in lb) 11. Order 1 consists of lb of cashews and lb of peanuts and it costs $16, Order consists of 7 lb of cashews and lb of peanuts and it costs $18. Organize this information into a table and solve foe x and y. 1. Order 1 consists of lb of cashews and lb of peanuts and it costs $8, Order consists of lb of cashews and lb of peanuts and it costs $0. Organize this information into a table and solve for x and y. Note that 9 1 are the same as the hamburger fries problem. C. Select the x and y, make a table, then solve. 1. We have $000 more saved in an account that gives % interest than in an account that gives % interest. In one year, their combined interest is $600. How much is in each account? 1. We have saved in a 6% account $000 less than twice than in a % account. In one year, their combined interest is $1680. How much is in each account? 1. The combined saving in two accounts is $1,000. One account has % interest, the other has % interest. The combined interest is $70 in one year. How much is in each account? 16. The combined saving in two accounts is $1,000. One account has 6% interest, the other has % interest. The combined interest is $70 in one year. How much is in each account? 6

65 Chapter Polynomial Expressions and Operations 1 Exponents We write the quantity A multiplied to itself N times as A N, i.e. A x A x A.x A = A N exponent Example A. base = ()()() = 6 (xy) = (xy)(xy) = x y xy = (x)(yy) x = (xx) Rules of Exponents Multiplication Rule: A N A K =A N+K Example B. a. = (**)(***) = + = 7 b. x y 7 x y 6 = x x y 7 y 6 = x 9 y 1 Division Rule: A N A K = A N K Example C. = ()()()()()() = ()() = Power Rule: (A N ) K = A NK Example D. ( ) = ( )( )( )( )( ) = ++++ = * = 0 Since A 1 A 1 = 1 = A 1 1 = A 0, we obtain the 0-power Rule. 0-Power Rule: A 0 = 1, A = 0 Since 1 = A 0 A K = A 0 K = A K, we get the negative-power Rule. A K Negative-Power Rule: A K = 1 A K Example D. Simplify a. 0 = 1 b. = 1 1 = 9 c. ( ) 1 = / 1 = 1* = a In general ( b ) K b = ( a ) K d. ( ) = ( ) =, A = 0 1 e. 1 0 * = 1 1* = 1 1 = 1 1 Although the negative power means to reciprocate, for problems of consolidating exponents, we do not reciprocate the negative exponents. Instead we add or subtract them using the multiplication and division rules first. Example E. Simplify x y 6 x 8 y x y 6 x 8 y = x x 8 y 6 y = 1 x 8 y = x y 9 17 = 1 y 17 9x = y 17 9x 6

66 Example F. Simplify x 8 using the rules for exponents. 6 x Leave the answer in positive exponents only. x 8 = 6 x 8 ( ) 6 x = x = 1 1 * = 1 8x x Example G. Simplify (x y ) x x (y 1 x ) (x y ) x = x y 6 x x y x 6 = x x y 6 x (y 1 x ) x x 6 y = x 6 y 6 x y = ( ) x 6 y = x y = 7 x y 6 ( ) Exponents Exercise. A. Write the numbers without the negative exponents and compute the answers ( ) ( 1 ) 9. ( ) * * 1 0 * 1 1. * * 1. * * 1 ( ) B. Combine the exponents. Leave the answers in positive exponents but do not reciprocate the negative exponents until the final step. 16. x x 17. x x 18. x x 19. x x 0. x y x y 1. y x y x. x xy x. y 1 x y x 9. x y x 1 y 11. x (x ) 6. (x ) x 6 7. x (x y ) y 8 B. Combine the exponents. Leave the answers in positive exponents but do not reciprocate the negative exponents until the final step x 8 x x 6 x y 8 y x y 1 (x y ) (y x ) 1 9. x 8 x. x 6 y 8 y x x 8 x. 6 y (x y ) (x y ) y C. Combine the exponents as far as possible. 1. y x x y y 6 x 8 x y 8. x 9. x+ x 0. a x a x+ 1. (b ) x+1 b x+. e e x+1 e x. e e x+1 e x. How would you make sense of? 66

67 Scientific Notation An important application for exponents is the use of the powers of 10 in calculation of very large or very small numbers. Powers of 10: 10 = = = = = = = = Scientific Notation Any number can be written in the form A x 10 N where 1 < A < 10. This form is called the scientific notation of the number. To write a number in scientific notation, we move the decimal point behind the first nonzero digit. i. If the decimal point moves to the left N spaces, then the exponent over 10 is positive N, i.e. if after moving the decimal point we get a smaller number A, then N is positive. ii. If the decimal point moves to the right N spaces, then the exponent over 10 is negative, i.e. if after moving the decimal point we get a larger number A, then N is negative. Example A. Write the following numbers in scientific notation. a = = 1. x 10 + Move left places. b = = 1. x 10 Move right places To change a number in scientific notation back to the standard form, we move the decimal point according to N. i. If N is positive, move the decimal point in A to the right, i.e. make A into a larger number. ii. If N is negative, move the decimal point in A to the left, i.e. make A into a smaller number. Example B. Write the following numbers in the standard form. a. 1. x 10 + = = 100. Move right places, b. 1. x 10 = = Move left places Scientific notation simplifies multiplication and division of very large and very small numbers. 67

68 Example C. Calculate. Give the answer in both scientific notation and the standard notation. a. (1. x 10 8 ) x (1. x 10 1 ) = 1. x 1. x 10 8 x 10 1 = 1.6 x = 1.6 x 10 = b. 6. x x = 6. x 10 ( 10).1 = x 10 8 = 00,000,000 Example D. Convert each numbers into scientific notation. Calculate the result. Give the answer in both scientific notation and the standard notation. 0,000,000 x =. x 10 8 x. x x 10 =. x. x 108 x x 10 =. x. 1. x ( 6) ( ) = x = x 10 6 =,000,000 68

69 Scientific Notation Ex.ercise A. Convert the following numbers into standard notation. Remember that multiplying by 10 positive power makes a number larger, and multiplying by 10 negative power makes the number smaller x x x x x x x x 10 6 B. Convert the following numbers in scientific notation into standard notation x x x x x x x x C. Convert the following numbers into the scientific notation , ,000. 1,000,000,000, C. Convert the following numbers into the scientific notation. 8., D. Calculate by converting into the scientific notation first. Put the answers in both the scientific and standard notations x x x ,000 x 1,00, x ,000 x x 1,00, x,00, x 600, x ,000,000 x 600,000, x x 600, x ,00,000,000 x 6, x x Google scientific notation applications and list one that impresses you. 69

70 Polynomial Expressions A mathematics expression is a calculation procedure written in numbers, variables, and operation symbols. Example A. + x the sum of and times x x x the difference between times the square of x and times x ( x) the square of the difference of and twice x An expression of the form #x N, where the exponent N is a non-negative integer and # is a number, is called a monomial (one-term). For example, x, x, and x 6 are monomials. Example B. Evaluate the monomials if y = a. y y ( ) = (16) = 8 b. y (y = ) y ( ) = (16) = 8. c. y y ( ) = ( 6) = 19 Polynomial Expressions The sum of monomials are called polynomials (many-terms), these are expressions of the form #x N ±#x N-1 ± ±#x 1 ±# where # can be any number. For example, x + 7, x x + 7, x + 1 are polynomials, whereas the expression 1 x is not a polynomial. Example C. Evaluate the polynomial x x if x =. The polynomial x x is the combination of two monomials; x and x. When evaluating the polynomial, we evaluate each monomial then combine the results. Set x = ( ) in the expression, we get ( ) ( ) = (9) ( 7) = = 117 Given a polynomial, each monomial is called a term. #x N ±#x N-1 ± ±#x ±# terms Therefore the polynomial x x + 7 has terms, x, x and

71 Each term is addressed by the variable part. Hence the x -term of the x x + 7 is x, the x-term is x, and the number term or the constant term is 7. The number in front of a term is called the coefficient of that term. So the coefficient of x is. Operations with Polynomials Terms with the same variable part are called like-terms. Like-terms may be combined. For example, x + x = 9x and x x = x. Unlike terms may not be combined. So x + x stays as x + x. Note that we write 1x N as x N, 1x N as x N. When multiplying a number with a term, we multiply it with the coefficient. Hence, (x) = (*)x =1x, and ( x) = ( )( )x = 8x. When multiplying a number with a polynomial, we may expand using the distributive law: A(B ±C) = AB ±AC. Example D. Expand and simplify. a. (x ) + ( x) = 6x x = x b. (x x + ) ( x x 6) = x + 9x 1 + x + 8x +1 = x + 17x Polynomials in Two or More Variables We form polynomials in two variables say, x & y, by adding monomials of the form kx # y # where k is a number and the powers are all nonnegative integers such as x y or x. Like terms are terms where the variable parts are the same. For example x y + x y = 8x y but x y + x y can t be combined. We evaluate them by assigning numbers to x and/or y. If only one number is given, the result is a formula. If both numbers are given, then we get a numerical output. We may do this for x, y and z or even more variables. Example E. Expand and simplify. a. (xy x y) + xy xy = 6xy 8x y + xy xy = 8xy 8x y xy b. Evaluate 8xy 8x y xy if x =. Input x =, we get 8()y 8() y ()y = 16y y 6y = 16y 6y c. Evaluate 8xy 8x y xy if x = and y = We may put x =, y = into the formula and do everything all over again or we may plug into y = into part b which is easier. We will do the easy way. Input y = into 16y 6y we get 16() 6() = 8 = 10 71

72 Polynomial Expressions Exercise. A. Evaluate each monomials with the given values. 1. x with x = 1 and x = 1. x with x = 1 and x = 1. x with x = 1 and x = 1. x with x = 1 and x = 1. y with y = and y = 6. y with y = and y = 7. z with z = and z = 8. y with z = and z = B. Evaluate each monomials with the given values. 9. x x + with x = 1 and x = x + x 1 with x = and x = 11. x x with x = and x = 1. x x + with x = and x = 1. x x + with x = and x = 1. x x with x = and x = C. Expand and simplify. 1. (x ) + ( x) 16. (x ) ( x) 17. (x 8) + ( 9x) 18. (x 8) ( 9x) 19. 7( x 7) ( x) 0. ( 8x) + 7( 11x) 1. x x + + ( x x 6). x x + ( x x 6). (x x + ) + ( x x 6). (x x + ) ( x x 6). (x x + ) + ( x x ) 6. (x x + ) ( x x ) 7. (x x ) 9(6x 7x) ( 8x ) 8. 6(7x + x 9) 7( x x 7) 9. Simplify (xy xy ) (xy xy ) then evaluated it with x = 1, afterwards evaluate it at ( 1, ) for (x, y) 0. Simplify x (xy x ) (y xy) then evaluated it with y =, afterwards evaluate it at ( 1, ) for (x, y) 1. Simplify x (xy z ) (z x ) then evaluated it with x = 1, y = and z =. 7

73 Polynomial Operations In the last section, we introduced polynomial expressions. Given a polynomial, each monomial is also called a term. #x N ±#x N-1 ± ±#x ±# terms Terms with the same variabe part are called like-terms. Like-terms may be combined. For example, x + x x x = 9x x. Unlike terms may not be combined. In particular x + x = x = x, and that x + x = x + x = x. Note that we write 1x N as x N, 1x N as x N. When multiplying a number with a term, we multiply it to the coefficient. Hence, (x) = (*)x =1x, and ( x) = ( )( )x = 8x. When multiplying a number to a polynomial, we may expand the result using the distributive law: A(B ±C) = AB ±AC. Example A. Expand and simplify. a. (x ) + ( x) = 6x x = x b. (x x + ) ( x x 6) = x + 9x 1 + x + 8x +1 = x + 17x When multiply a term with another term, we multiply the coefficient with the coefficient and the variable with the variable. Example B. a. (x )(x ) = *x x = 6x b. x ( x) = ( )x x = 1x c. x (x x) distribute = 6x 1x To multiply two polynomials, we may multiply each term of one polynomial against other polynomial then expand and simplify. Example C. a. (x + )(x 1) = x(x 1) + (x 1) = 6x x + x = 6x + x b. (x 1)(x + x ) = x(x + x ) 1(x + x ) = x + 6x 8x x x + = x + x 11x + Note that if we did (x 1)(x + ) or (x + x )(x 1) instead, we get the same answers. (Check this.) Fact. If P and Q are two polynomials then PQ QP. A shorter way to multiply is to bypass the nd step and use the general distributive law. 7

74 General Distributive Rule: (A ±B ±C ±..)(a ±b ±c..) = Aa ±Ab ±Ac..±Ba ±Bb ±Bc..±Ca ±Cb ±Cc.. Example D. Expand a. (x + )(x ) = x x + x 1 simplify = x x 1 b. (x )(x x ) = x x x x + 6x + 6 = x x + x + 6 We will address the division operation of polynomials laterafter we understand more about the multiplication operation. Exercise. A. Multiply the following monomials. 1. x ( x ). x (8x ). x ( x ) 11. x(x ) 9(6x 7) Polynomial Operations. 1( x 6 ). ( 8 x ) 7. 1x ( x ) B. Fill in the degrees of the products. 8. #x(#x + # x + #) = #x? + #x? + #x? 9. #x (#x + # x + #x ) = #x? + #x? + #x? C. Expand and simplify. 6. 6x ( x ) 10. #x (#x + # x + #x + #) = #x? + #x? + #x? + #x? 1. x(x + 7) + (x ) 1. x(x + ) 8x(7x ) 1. x( x + 9) + 6x(6x 1) 1. x( x + ) x(x 1) (x ) 16. x( 7x + 9) x(x ) + 9(x + ) D. Expand and simplify. (Use any method.) 18. (x + )(x + 7) 19. (x )(x + 7) 0. (x + )(x 7) 1. (x )(x 7). (x )(x + ). ( x + )(x + 8). (x )(x + ). (x + 7)(x 7) 6. (x )(x 6) 7. (8x )( x 7) 8. (x 7)(x x + 9) 9. (x + )(x x + ) 0. (x 1)(x 1) 1. (x + 1). (x ). (x + ). x(x 1)(x + ). x(x )(x + ) 6. (x )(x 1)(x + ) 7. (x + 1)(x + 1)(x ) 8. (x 1)(x + 1) 9. (x 1)(x + x + 1) 0. (x 1)(x + x + x + 1) 1. (x 1)(x + x + x + x + 1). What do you think the answer is for (x 1)(x 0 + x x + x + 1)? 7

75 Special Binomial Operations A binomial is a two-term polynomial. Usually we use the term for expressions of the form ax + b. A trinomial is a three term polynomial. Usually we use the term for expressions of the form ax + bx + c. The product of two binomials is a trinomial. (#x + #)(#x + #) = #x + #x + # F: To get the x -term, multiply the two Front x-terms of the binomials. OI: To get the x-term, multiply the Outer and Inner pairs and combine the results. L: To get the constant term, multiply the two Last constant terms. This is called the FOIL method. The FOIL method speeds up the multiplication of above binomial products and this will come in handy later. Example A. Multiply using FOIL method. a. (x + )(x ) = x x 1 The last terms: 1 b. (x + )( x + ) = 6x + 7x + 0 The last terms: 0 Expanding the negative of the binomial product requires extra care. One way to do this is to insert a set of [ ] around the product. Example B. Expand. a. [(x )(x + )] Insert [ ] = [ x + 1x x 0] Expand = [ x + 11x 0] = x 11x + 0 Remove [ ] and change signs. Another way to do this is to distribute the negative sign into the first binomial then FOIL. Example C. Expand. a. (x )(x + ) = ( x + )(x + ) = x 1x + x + 0 = x 11x + 0 Distribute the sign. Expand b. Expand and simplify. (Two versions) (x )(x +) (x )(x + ) = (x )(x +) + ( x + )(x + ) = x + 6x x 1 + ( x 1x + x + 0) = x + x 1 x 11x + 0 = x 10x + (x )(x +) [(x )(x + )] Insert brackets = x + x 1 [x +11x 0] = x + x 1 x 11x + 0 = x 10x + Expand Remove brackets and combine 7

76 If the binomials are in x and y, then the products consist of the x, xy and y terms. That is, (#x + #y)(#x + #y) = #x + #xy + #y The FOIL method is still applicable in this case. Example D. Expand. (x y)(x + y) = x + 1xy yx 0y = x + 11xy 0y Special Binomial Operations Exercise. A. Expand by FOIL method first. Then do them by inspection. 1. (x + )(x + 7). (x )(x + 7). (x + )(x 7). (x )(x 7). (x )(x + ) 6. ( x + )(x + 8) 7. (x )(x + ) 8. (x + 7)(x 7) 9. ( x + 7)(x + ) 10. ( x + )(x ) 11. (x )(x + ) 1. (x + 7)(x 7) 1. (9x + )(x ) 1. ( x + )( x + 1) 1. (x 1)(x ) 16. (6x )( x + 7) 17. (x + y)(x 7y) 18. (x y)(x 7y) 19. (x + 7y)(x 7y) 0. ( x + y)( x + y) B. Expand and simplify. 1. (x )(x + ). (6x 1)(x ). (8x )(x + 1). (x )(x ) C. Expand and simplify.. (x )(x + ) + (x )(x + ) 6. (x 1)(x ) + (x + )(x + ) 7. (x )(x + ) + (x + )(x ) 8. (x )(x + ) (x )(x + ) 9. (x )(x ) (x + )(x + ) 0. (x )(x + ) (x + )(x ) 1. (x 7)(x ) (x 1)(x + ). (x 1)(x 7) (x 7)(x + 1). (x )(x + ) (x + )(6x ). (x ) (x 1). (x 7) (x + ) 6. (x + ) (6x ) 76

77 6 Multiplication Formulas There are some important patterns in multiplying expressions that it is worthwhile to memorize. The two binomials (A + B) and (A B) are said to be the conjugate of each other. For example, the conjugate of (x + ) is (x ), and the conjugate of (ab c) is (ab + c). Note: The conjugate is different from the opposite. The opposite of (x + ) is ( x ). I. Difference of Squares Formula (A + B)(A B) = A B Conjugate Product To verify this : (A + B)(A B) = A AB + AB B = A B Difference of Squares Here are some examples of squaring: (x) = 9x, (xy) = x y, and (z ) = z. Example A. Expand. a. (x + )(x ) = (x) () = 9x (A + B)(A B) = A B b. (xy z )(xy + z ) = (xy) (z ) = x y z II. Square Formulas (A + B) = A + AB + B (A B) = A AB + B We may check this easily by multiplying, (A + B) = (A + B)(A + B) = A + AB + BA + B = A + AB + B We say that (A + B) is A, B, plus twice A*B, and (A B) is A, B, minus twice A*B. Example B. a. (x + ) = (x) + (x)() + = 9x + x + 16 (A + B) = A + AB + B b. (a b) = (a) (a)(b) + (b) = 9a 0ab + b III. Some Applications of the Formulas We can use the above formulas to help us multiply. Example C. Calculate. Use the conjugate formula. a. 1*9 = (0 + 1)(0 1) = 0 1 =,00 1 =,99 b. *8 = (0 + )(0 ) = 0 =,00 =,96 c. 6*7 = (60 + )(60 ) = 60 =,600 9 =,91 The conjugate formula (A + B)(A B) = A B may be used to multiply two numbers of the forms (A + B) and (A B) where A and B can be calculated easily. 77

78 The Squaring Formulas. (A + B) is A, B, plus twice A*B, (A B) is A, B, minus twice A*B. Example D. Calculate. Use the squaring formulas. a. 1 = (0 + 1) = (0)(1) =, =,601 b. 9 = (0 1) = (0)(1) =, =,01 b. (0½) = (0 + ½ ) = 0 + ½ + (½) (0) =,00 + 1/ + 0 =,0¼ Multiplication Formulas Exercise. A. Calculate. Use the conjugate formula. 1. 1*19. 1*9. 1*9. 71*69. *18 6. *8 7. *8 8. 7*67 B. Calculate. Use the squaring formula ½ ½ 16. 9½ C. Expand. 18. (x + )(x ) 19. (x 7)(x + 7) 0. (x + )(x ) 1. (x )(x + ). (7x + )(7x ). ( 7 + x )( 7 x). ( x + )( x ). (x y)(x + y) 6. (x y)(x + y) 7. (1 7y)(1 + 7y) 8. ( x)( + x) 9. (10 + 9x)(10 9x) 0. (x + ) 1. (x 7). (x + ). (x + y). (7x y). (x h) 7 Polynomial Operations II In this section, we carry out the operations in a vertical format. To add and subtract in the vertical format, line up like-terms vertically then add or subtract. Example A. a. Do vertically ( x + ) + (6x 8) x + +) 6x 8 x Hence ( x + ) + (6x 8) = x b. Do vertically (x + 8x + ) + ( 6x 8x ) x + 8x + +) 6x 8x x Hence (x + 8x + ) + ( 6x 8x ) = x + 78

79 Vertical subtraction is used often. For examples, x ) 6x x ) 6x ) x 6x x ) 6x x 10x 10x x One way to help the subtraction is to change the signs of terms subtracted then add. Example B. a. Do vertically ( x + ) (6x 8) x + ) 6x + 8 change signs and add 10x +1 Hence ( x + ) (6x 8) = 10x + 1 Example C. Do vertically (x + ) ( 6x + 8x ) x ) + 6x + 8x + 10x 8x + 8 Hence (x + ) ( 6x + 8x ) = 10x 8x + 8 To multiply vertically, line up the like-terms in the product. Example D. Do vertically ( x + )(x ) x + x) x 8x + 1x +) + 10x 1 8x + x 1 Hence ( x + )(x ) = 8x +x 1 We may switch the order of multiplication. Example D. Do vertically (x x )(x + ) x x x) x + x 1x 9x +) x 8x 6 x 10x 17x 6 Hence (x x )(x + ) = x 10x 17x 6 Exercise. Do problem 18 9 from the last section using the vertical format. 79

80 Chapter Factoring and Polynomial Equations To factor means to rewrite a quantity as a product (in a nontrivial way). A quantity x that can t be written as product besides as 1*x is said to be prime. To factor completely means each factor in the product is prime. Example A. Factor 1 completely. 1 = * = * * 1 Factoring Out GCF not prime factored completely A common factor of two or more quantities is a factor belongs to all the quantities. Example B. a. Sincer 6 = *, 1 = *, is a common factor. b. The common factor of ab, 6a are, a, a. c. The common factors of 6xy, 1x y are, x, y, xy,.. d. The common factor of a(x+y), b(x+y) is (x+y). The greatest common factor (GCF) is the common factor that has the largest coefficient and highest degree of each factor among all common factors. Example C. Find the GCF of the given quantities. a. GCF{, 6} = 1. b. GCF{ab, 6a} = a. c. GCF {6xy, 1 x y } = xy. d. GCF{x y, x y 6, x y } = x y. The Extraction Law Distributive law interpreted backward gives the Extraction Law, that is, common factors may be extracted from sums or differences. AB AC A(B C) This procedure is also called factoring out a common factor. To factor, the first step always is to factor out the GCF. Example D. Factor out the GCF. a. xy y = y(x ) (the GCF is y) b. ab + 6a = a(b) + a() = a(b + ) (the GCF is a) c. 1x y + 6x y = 6x y (y) + 6x y (1) = 6x y (y + 1) (the GCF is 6x y ) We may pull out common factors that are ( )'s. Example E. Factor a. a(x + y) (x + y) Pull out the common factor (x + y) a(x + y) (x + y) = (x + y)(a ) b. Factor (x )x (x ) Pull out the common factor (x ), (x )x (x ) = (x )(x ) 80

81 We sometimes pull out a negative sign from an expression. Example F. Pull out the negative sign from x +. x + = (x ) Sometime we have to use factor twice. Example G. Factor by pulling out twice. a. y(x ) x + = y(x ) (x ) = (x )(y 1) b. x y + ax ay Group them into two groups. = (x y) + (ax ay) Factor out the GCF of each group. = (x y) + a(x y) Pull the factor (x y) again. = ( + a)(x y) Factoring Out GCF Exercise. A. Find the GCF of the listed quantities. 1. {, 6 }. {1, 18 }. {, 0, 1 }. {, 0, 0 }. {x, 6x } 6. {1x y, 18xy } 7. {A B, 0A B, 1 A B } 8. {x 7 y 6 z 6, 0y 7 z x 6, 0z 8 x 7 y 6 } B. Factor out the GCF. 9. 6y 10. 1x + 18y 11. A + 0B 1C 1. x + 0y 0 1. x + 6x 1. 1x y 18xy 1. A B 0A B 1A B } 16. x 7 y 6 z 6 0y 7 z x 6 + 0z 8 x 7 y x 8x + x 18. 0x x 19. x(x ) + (x ) 0. x(x ) (x ) C. Factor out the. 1. y +. x x x + 16 D. Factor, use grouping if it s necessary.. y y + y 6 6. x + x + 6x y y y x + x 6x y y + y 0. x x x y y 6y +. x + x 6x. x + 6x 6x 9. x + x 6x + 8. y + 10y y x + x 7x y xy 6xy + x 8. x + xy 6xy y 9. x + xy 0xy + 8y 0. 1x + 1xy 8xy + 1y 81

82 Factoring Trinomials I Trinomials (three-term) are polynomials of the form ax + bx + c where a, b, and c are numbers. The product of two binomials is a trinomials: (#x + #)(#x + #) ax + bx + c Hence, to factor a trinomial, we write the trinomial as a product of two binomials, if possible, that is: ax + bx + c (#x + #)(#x + #) We start with the case where a = 1, or trinomials of the form x + bx + c. To factor x + bx + c, we note hat if (x + u)(x + v) = x + ux + vx + uv = x + (u + v)x + uv = x + bx + c, we need to have u and v where uv = c, and u + v = b. If this can t be done, then the trinomial is prime (not factorable). Example A. a. Factor x + x + 6 We want (x + u)(x + v) = x + x + 6, so we need u and v where uv = 6 and u + v =. Since 6 = (1)(6) = ()() = (-1)(-6) = (-)(-) and + =, so x + x + 6 = (x + )(x + ). b. Factor x x + 6 We want (x + u)(x + v) = x x + 6, so we need u and v where uv = 6 and u + v =. Since 6 = (1)(6) = ()() = (-1)(-6) = (-)(-) and =, so x x + 6 = (x )(x ). c. Factor x + x 6 We want (x + u)(x + v) = x + x 6, so we need uv = 6 and u + v =. Since -6 = ( 1)(6) = (1)( 6) = ( )() =()( ) and =, so x + x 6 = (x 1)(x + 6). Observations About Signs Given that x + bx + c = (x + u)(x + v) so that uv = c, we observe the following. 1. If c is positive, then u and v have same sign. In particular, if b is also positive, then both are positive. { if b is negative, then both are negative. From the examples above x + x + 6 = (x + )(x + ) x x + 6 = (x )(x ). If c is negative, then u and v have opposite signs. The one with larger absolute value has the same sign as b. From the example above x x 6 = (x 6)(x + 1) 8

83 Example B. a. Factor x + x 1 We need u and v having opposite signs such that uv = 1, u + v = +. Since -1 = (-1)(1) = (-)(6) = (-)() They must be and 6 hence x + x 1 = (x )(x + 6). b. Factor x 8x 1 We need u and v such that uv = 1, u + v = 8 with u and v having opposite signs. This is impossible. Hence x 8x 1 is prime. Factoring Trinomials I Exercise. A. Factor. If it s prime, state so. 1. x x. x + x. x x 6. x + x 6. x x + 6. x + x 7. x + x 8 8. x x 9. x + x x + x 6 1. x x x x 0 1. x 9x x 9x x x 6 1. x 8x x 10x 18. x 10x x 11x + 0. x 11x 1. x 1x 6. x 1x + 6. x 1x 6. x 1x + 6 B. Factor. Factor out the GCF, the, and arrange the terms in order first if necessary. 7. x x x 18x + 0x 9. x 0x 7 0. x + 0x x. x 6 6. x x + 18x. x x + x. x + 10x x 0. x xy + 96y Factoring Trinomials I C. Factor. Factor out the GCF, the, and arrange the terms in order first.. yx + yx + y. x 0x 8x 6. x + 0x x 7. x + 11xy + y 8. x 6x + 6x 9. x + 9xy + 6y D. Factor. If not possible, state so. 1. x + 1. x +. x + 9. x +. What can you conclude from 1? 8

84 Factoring Trinomials II Now let s try to factor trinomials of the form ax + bx + c. We ll give two methods. One is short but not reliable. The second one takes more steps but gives definite answers. Reversed FOIL Method For this method, we need to find four numbers that fit certain descriptions. The following are examples of the task to be accomplished. Example A. Let {1, } and {1, } be two pairs of numbers. Is it possible to split the {1, }, put them in the boxes that makes the equality true? a. 1* (± ) + *(± ) =. Yes, 1* () + * (1) = b. 1* (± ) + * (± ) =. Yes, 1* (1) + * ( ) = or 1* ( ) + * ( 1) = c. 1* (± ) + * (± ) = 8. No, since the most we can obtain is 1* (1) + * () = 7. (Reversed FOIL Method) Let s see how the above examples are related to factoring. Example B. Factor x + x +. The only way to get x is (x ±#)(1x ±#). The # s must be 1 and to get the constant term +. We need to place 1 and as the #'s so the product will yield the correct middle term +x. That is, (x ±#)(1x ±#) must yields +x, or that (±# ) +1(±#) = where the # s are 1 and. Since (1) +1() =, we see that x + x + = (x + )(1x + 1). x Example C. Factor x 7x +. We start with (x ±#)(1x ±#). We need to fill in 1 and as #'s so that (±# ) + 1(±# ) = 7. It's ( ) + 1( 1) = 7. So x 7x + = (x 1)(1x ) Example D. Factor x + x. We start with (x ±#)(1x ±#). We need to fill in 1 and so that (±# ) + 1(±# ) = +. Since c is negative, they must have opposite signs. It is (+) + 1( 1) = +. So x + x + = (x 1)(1x + ) 8

85 Example E. Factor x + 8x +. We start with (x ±#)(1x ±#). We need to fill in 1& so that (±# ) + 1(±# ) = +8. This is impossible. Hence the expression is prime. If both the numbers a and c in ax + bx + c have many factors then there are many possibilities to check. Example F. Factor x + 11x. We start with (x ±#)(1x ±#). Since = () = 1(), we need to fill in & or 1& so that (±# ) + 1(±# ) = +11. It can't be &. Try 1&, it is (+) + 1( 1) = +11. So x + 11x = (x 1)(1x + ). It's not necessary to always start with ax. If c is a prime number, we start with c. Example G. Factor 1x x. Since must be (1), we need to find two numbers such that (#)(#) = 1 and that (±#)(±) + (±#)(±1) =. 1 = 1(1) = (6) = () 1&1 and &6 can be quickly eliminated. We get ()( ) + ()(+1) =. So 1x x = (x + 1)(x ). Remark: In the above method, finding (#)(±#) + (#)( ±#) = b does not guarantee that the trinomial will factor. We have to match the sign of c also. Example H. Factor x 7x. We start with (x ±#)(1x ±#). We find that: ( ) + 1( 1) = 7. But this won't work since ( )( 1) = = c. In fact this trinomial is prime. There might be multiple matchings for (#)(±#) + (#)( ±#) = b make sure you chose the correct one, if any. Example I: Factor 1x + x 6. We have: 1(+) + 1(+) = + 1(+6) + 1( 1) = + The one that works is x + x 6 = (x + 6)(x 1). 8

86 Finally, before starting the reverse-foil procedure 1. make sure the terms are arranged in order.. if there is any common factor, pull out the GCF first.. make sure that x is positive, if not, factor out the negative sign first. Example J. Factor x + x + x x + x + x Arrange the terms in order = x + x + x Factor out the GCF = x(x x ) = x(x )(x + 1) Factoring Trinomials II Exercise. A. Factor completely. If it s prime, state so. 1. x x. x + x. x x 1. x + x 1. x x x + x 1 7. x + x 8. x x 9. x x 10. x + 9x 11. x + x + 1. x x + 1. x x x x x + x x x 17. 6x 1x x 1x x + 7x x 7x x 1x x + 1x x x. 6x 1x x 1x 8. x 9 6. x 9 7. x 8. x x + 9 B. Factor. Factor out the GCF, the, and arrange the terms in order first. 0. 6x xy + 6y 1. x + x x. 6x x + x. 1x x 10x. 1x y 1x y + xy 86

87 Factoring Trinomials III In this section we give a formula that enables us to tell if a trinomial is factorable or not. Also we give the ac-method for factoring. Theorem: If b ac = 0, 1,, 9, 16,, 6,.. i.e. is a squared number, then the trinomial is factorable. Otherwise, it is not factorable. Example I. Use the b ac to see if the trinomial is factorable. If it is, factor it. a. x 7x b ac = ( 7) ()( ) = 9 + = 7 is not a square, hence it is prime. b. x 7x + b ac = ( 7) ()() = 9 = which is a squared number, hence it is factorable. In fact x 7x + = (x 1)(x ) The ac-method The product of two binomials has four terms. We may use grouping-method to work backward to obtain the two binomials. Example A. Factor x x 6 by grouping. x x 6 = x x + x 6 Put them into two groups = (x x) + (x 6) Take out the common factors = x(x ) + (x ) Take out the common (x ) = (x )(x + ) But how do we know to write x x 6 as x x + x 6 in the first place? This leads to the ac-method. The ac-method tells us how to rewrite the trinomial so we may use the grouping method. If it isn't possible to rewrite the trinomial according to the ac-method, then the trinomial is prime. 87

88 ac-method: We assume that there is no common factor for the trinomial ax + bx + c. 1. Calculate ac, and find two numbers u and v such that uv is ac, and u + v = b.. Write ax + bx + c as ax + ux + vx +c. Use the grouping method to factor (ax + ux) + (vx + c) If step 1 can t be done, then the expression is prime. Example B. Factor x x 0 using the ac-method. a =, c = 0. Hence ac = ( 0) = 60. We need two numbers u and v such that uv = 60 and u + v =. By searching, we get that 6( 10) = 60 and 6 + ( 10) =. Write x x 0 = x + 6x 10x 0 Put in two groups, = (x + 6x ) + ( 10x 0) Pull out common factor = x(x + ) 10 (x + ) Pull out common factor = (x 10)(x + ) Example C. Factor x 6x 0 using the ac-method. a =, c = 0, hence ac = (-0) = 60. Need two numbers u and v such that uv = 60 and u + v = 6. After searching all possibilities we found that it's impossible. Hence x 6x 0 is prime. Factoring Trinomials III Exercise A. Use the ac method, factor the trinomial or demonstrate that it s not factorable. 1. x x. x + x. x x 1. x + x 1. x x x + x 1 7. x + x 8. x x 9. x x 10. x + 9x 11. x + x + 1. x x + 1. x x x x x + x x x 17. 6x 1x x 1x x + 7x x 7x x 1x x + 1x x x. 6x 1x x 1x 8. x 9 6. x 9 7. x 8. x x + 9 B. Factor. Factor out the GCF, the, and arrange the terms in order first. 0. 6x xy + 6y 1. x + x x. 6x x + x. 1x x 10x. 1x y 1x y + xy 88

89 Substitution and Factoring Formulas We may evaluate expressions with numbers as inputs. For example, evaluate A B with A = and B = 1 as inputs, we have. We extend this to evaluating expressions with symbols as inputs. This is called "substitution". In such problems always simplify the answers as much as possible. Example A. Evaluate A B with A = (x + ) and B = (x). Replace A by (x + ), and B by (x) we have (x + ) (x) = x + x + x = x + x + Vice versa, for some outputs we can easily identify the inputs especially when the formula is given. Example B. With inputs for A and B, evaluate A B a. we have x 9y, what are A and B? Match the A = x, we re asking (? ) = x, so A = x. Similarly we conclude that B = y. b. we have x y 1, what are A and B? Match the A = x y and B = 1, we have that A = xy and B = 1. c. we have x + xy + y 9, what are A and B? Factor x + xy + y we have (x + y)(x + y) = (x + y), so x + xy + y 9 = (x + y). Hence A = (x + y) and B =. Example C. Evaluate A B with inputs A and B a. we have x 1, what are A and B? Match the A = x, we have A = x. Similarly we ve B = 1 so B =. b. we have 8y ( y), what are A and B? Match the A =8y, we have A = y. By inspection, we have B = ( y). Recall the conjugate product formula (A B)(A + B) = A B. The reverse of the conjugate product formula is the factoring formula for difference of squares : A B = (A B) (A + B) As suggested by the name, we use this formula when it is the subtraction of two perfect square terms. Example D. a. x = (x) () = (x + )(x ) b. x 9y = (x) (y) = (x y)(x + y) c. A B 1 = (AB ) (1) = (AB 1)(AB + 1) 89

90 d. (A + B) 16 = (A + B) () = ((A + B) )((A + B) + ) = (A + B )(A + B + ) Caution: The expressions A + B or A B are not factorable! Sum and Difference of Cubes Both the difference and the sum of cubes may be factored. If we try to factor A B, we might guess that A B = (A B)(A + B ) But this does not work because we have the inner and outer products BA and AB remaining in the expansion as shown above. But if we put in the adjustment +AB, then all the crossproducts with A and B are eliminated. That is (A B)(A + AB + B ) = A + A B + AB BA B A B = A B Hence A B = (A B)(A + AB + B ) Through a similar argument we have that A + B = (A + B)(A AB + B ) We write these two formulas together as A + B = (A + B)(A + AB + B ) Note that the signs are reversed for the adjustments in these formulas. Example E. a. 8x 7 (Note that both 8 and 7 are cubes) = (x) () = (x )((x) + (x)() +() ) = (x )(x + 6x + 9) b. 6A + 1 (both 6 and 1 are cubes) = (A) + () = (A + )((A) (A)() +() ) = (A + )(16A 0A + ) Summary on Factoring I. Always factor out the GCF first. II. Make sure the leading term is positive and all terms are arranged in order. III. If left with a trinomial, use reverse-foil or ac-method. IV. If it is a two-term expression, try factoring by formula. V. If it is a four term expression, try the grouping-method. 90

91 Example F. Factor completely. A. 0A A Take out the common factor = A(A 9) Difference of square = A(A )(A + ) B. 0A + A 10A Take out the common factor and -1 = A(A 9A + ) Substitution and Factoring Formulas Exercise A. Simplify with the given substitution. Simplify your answers when possible. For 1-8, Simplify A B with the given substitution. 1. A = x, B = 1. A =, B = y. A = xy, B = 1. A = x + y, B =. A = (x ), B = x 6. A = x, B = (x 1) 7. A = (x ), B = (x 1) 8. A = (x + 1), B = (x 1) For 9-1, simplify A B with the given substitution. 9. A = x, B = 10. A = x, B = 11. A = x, B = y 1. A = 1, B = xy 1. A = x, B = 1. A = 6x, B = y 1. Simplify x x + 10 with the substitution x = ( + h) 16. Simplify x + x with the substitution x = ( + h) B. Identify the substitution for A and B then factor the expression using the pattern A B = (A B)(A + B). 17. x y 19. x 9 0. y 1. x y. x y 1. 98x 8. 81x y 6. x + xy + y 7. 0x 18y 8. 6x x 9. 8x z 9z 0. x y y 1 C. Identify the substitution for A and B then factor the expression using the pattern A ± B = (A ± B)(A + AB + B ). 1. x 1. x 8. 7 y. x y y 7. x x 6 9. x

92 6 Applications of Factoring There are many applications of the factored forms of polynomials. Following are some of them: 1. to evaluate polynomials,. to determine the signs of the outputs,. most importantly, to solve polynomial-equations. Evaluating Polynomials Often it is easier to evaluate polynomials in the factored form. Example A. Evaluate x x if x = 7 a. without factoring. b. by factoring it first. We get 7 (7) = 9 1 = x x = (x )(x+1) We get (7 )(7 + 1) = (8) = Example B. Evaluate x x + x for x = -, -1, by factoring it first. x x + x = x(x x + ) = x(x 1)(x ) For x = -: (-)[(-) 1] [(-) ] = - [-] [-] = -0 For x = -1: (-1)[(-1) 1] [(-1) ] = -1 [-] [-] = -9 For x = : [() 1] [() ] = [] [1] = 1 Determine the signs of the outputs Often we only want to know the sign of the output, i.e. whether the output is positive or negative. It is easy to do this using the factored form. Example C. Determine the outcome is + or for x x if x = -/, -1/. Factor x x = (x )(x + 1) Hence for x = -/: (-/ )(-/ + 1) is ( )( ) = +. So the outcome is positive. And for x = -1/: (-1/ )(-1/ + 1) is ( )(+) =. So the outcome is negative. Solving Equations The most important application for factoring is to solve polynomial equations. These are equations of the form polynomial = polynomial To solve these equations, we use the following obvious fact. Fact: If A*B = 0, then either A = 0 or B = 0 For example, if x = 0, then x must be equal to 0. 9

93 Example D. a. If (x ) = 0, then (x ) = 0, so x must be. b. If (x + 1)(x ) = 0, then either (x + 1) = 0 or (x ) = 0, x = 1 or x = To solve polynomial equation, 1. set one side of the equation to be 0, move all the terms to the other side.. factor the polynomial,. get the answers. Example E. Solve for x a. x x = 0 Factor (x )(x + 1) = 0 Hence x = 0 or x + 1 = 0 x = or x = -1 b. x(x + 1) = x + (1 x) Expand x + x = x + x x + x = x + Set one side 0 x + x x = 0 x + x = 0 Factor (x + )(x 1) = 0 Get the answers x + = 0 x = - x = -/ or x 1 = 0 x = 1 c. 8x(x 1) = 10x Expand 8x 8x = 10x Set one side 0 8x 8x 10x = 0 8x 18x = 0 Factor x(x + )(x ) = 0 Get the answers x = 0 or x + = 0 or x = 0 x = - x = x = -/ x = / Applications of Factoring Exercise A. Use the factored form to evaluate the following expressions with the given input values. 1. x x, x =,,. x x 1, x = 1,, 7. x x 1, x = ½,, ½. x x, x =,,. x x x, x =,, 6 6. x x + x, x =,, B. Determine if the output is positive or negative using the factored form. 7. x x, x = ½, /, ½, ¼ 8. x + x + 8, x = ½, /, ½, ¼ 9. x x 8x, x = ½, /, ¼, 6¼, 10. x x x, x = ½, /, ¼, ¼, 11. x x, x = 1., 0.87,., x x, x =.90,.19, 1.,.01 9

94 0. x (x ) = x C. Solve the following equations. 1. x x = 0 1. x x 1 = 0 1. x + 7x + 1 = x x + 8 = x = x x 1 = x x = x(x ) = 1. x = (x + 1) 1. x =. 8x =. 7x 1 = 0. x(x ) + = x 6. x(x ) + x + 6 = x + x 7. x(x + ) + 9 = ( x) 8. x x, x =,, 9. x x 8x = 0 1. x = x. x = x. x = x. 7x = x x. = (x + )(x + 1) 6. (x 1) = (x + 1) 7. (x + 1) = x + (x 1) 8. (x + ) (x + 1) = x 9. (x + ) (x + ) = (x + 1) 7 nd-degree-equation Word Problems Many physics formulas are nd degree. If a stone is thrown straight up on Earth then h = -16t + vt where h = height in feet t = time in second v = upward speed in feet per second Example A. If a stone is thrown straight up at a speed of 6 ft per second, a. how high is it after 1 second? t = 1, v = 6, so h = -16(1) + 6(1) h = = 8 ft height = -16t + vt after t seconds b. How long will it take for it to fall back to the ground? To fall back to the ground means the height h is 0. Hence h = 0, v = 6, need to find t. 0 = -16 t + 6t 0 = -16t(t ) t = 0 or t = 0 or t = Therefore it takes seconds. c. What is the maximum height obtained? Since it takes seconds for the stone to fall back to the ground, at seconds it must reach the highest point. Hence t =, v = 6, need to find h. h = -16() + 6() = = 6 (ft) Therefore, the maximum height is 6 feet. 9

95 Formulas of area in mathematics also lead to nd degree equations. Area of a Rectangle Given a rectangle, let L L = length of a rectangle W = width of the rectangle, w A = LW the area A of the rectangle is A = LW. If L and W are in a given unit, then A is in unit. For example: If L = in, W = in. then A = * = 6 in in 1 in 1 in in 1 in 6 in Example B. The length of a rectangle is inches more than the width. The area is 1 in. Find the length and width. Let x = width, then the length = (x + ) LW = A, so (x + )x = 1 x + x = 1 x + x 1 = 0 (x + 7)(x ) = 0 x = - 7 or x = Therefore, the width is and the length is 7. Area of a Parallelogram A parallelogram is the area enclosed by two sets of parallel lines. If we H=height move the shaded part as shown, we get a rectangle. Hence the area A of the parallelogram is A = BH. B=base Example C. The area of the parallelogram shown is 7 ft. Find x. The area is (base)(height) or that x(x + ) = 7 x + x = 7 x x + x 7 = 0 Factor x + (x + 9)(x ) = 0 x = 9/, x = ft Area of a Triangle Given the base (B) and the height (H) of a triangle as shown. Take another copy and place it above the original one as shown. We obtain a parallelogram. H If A is the area of the triangle, then A = HB or A = BH. B 9

96 Example D. The base of a triangle is inches shorter than the twice of the height. The area is 10 in. Find the base and height. Let x = height, then the base = (x ) Hence, use the nd version of the formula we have 0 = (x ) x 0 = x x 0 = x x 0 0 = (x )(x + ) x = or x = -/ Example D. The base of a triangle is inches shorter than the twice of the height. The area is 10 in. Find the base and height. Let x = height, then the base = (x ) Hence, use the formula A = BH *10 = (x ) x 0 = x x 0 = x x 0 x 0 = (x )(x + ) x = or x = -/ x Therefore the height is in. and the base is in. 96

97 nd-degree-equation Word Problems Exercise A. Use the formula h = 16t + vt for the following problems. 1. A stone is thrown upward at a speed of v = 6 ft/sec, how long does it take for it s height to reach 8 ft?. A stone is thrown upward at a speed of v = 6 ft/sec, how long does it take for it s height to reach 8 ft?. A stone is thrown upward at a speed of v = 96 ft/sec, a. how long does it take for its height to reach 80 ft? Draw a picture. b. how long does it take for its height to reach the highest point? c. What is the maximum height it reached?. A stone is thrown upward at a speed of v = 18 ft/sec, a. how long does it take for its height to reach 160 ft? Draw a picture. How long does it take for its height to reach the highest point and what is the maximum height it reached? B. Given the following area measurements, find x ft x 1 ft (x 1) 7. x x + 1 ft x + 8. x 1 ft x (x + ) ft (x 1) 18 ft x (x + 1) x cm x x 1. x x 9cm x 18km x + 1 km (x + ) (x + ) km x + 1 x km x 1 x 97

98 Solutions to the Odd Problems The Basic Pg. A B ,,,,, 6,,,,, 19., 7,, 7,,,, C Fractions and Mixed Numbers Pg. A. /, /,/,/,/7,/,/,/ B Multiplication and Division of Fractions Pg. 7 A. 1. a.6 b.10 c.1 d.18. a.10 b.1 c. d.7. a. b.1 c.1 d. 7. a.7 b. c.1 d.1 9. a.1 b. c.6 d.0 B Adults Teens Children Males Female 9 1 Total ; 19/7 LCM and LCD Pg. 10 A. 1. a. b. 18 c.1 d. 0. a.7 b. 90 c.180. a.0 b.60 c a.7 b.10 c.0 9. a.60 b.70 B Addition and Subtraction of Fractions Pg. 1 A

99 B C Some Facts About Divisibility Pg. 16 A can t be divided by and 6. 0 can be divided by and can be divided by and can be divided by and can be divided by and , 180, can be divided by ,180 can be divided by 18 Cross Multiplication Pg. 18 A B. 11. is more 1. 6 is more is more is more C Signed Numbers Pg. 0 A B Addition and Subtraction of Signed Numbers Pg. A B C D Multiplication and Division of Signed Numbers Pg. A B C Order of Operations Pg. 6 A B C.. and and D E

100 / Variables and Evaluation Pg / Expressions Pg. A. 1. 8x. -8x. 8x x + 9. x B-A B. 1. x x x x+ 1. -x+76. 6x+68. B-A 7. 9A-0B C. 9. 1x x. 0x x x/1 9. 9x/ 1. 0 bananas and 6 apples Linear Equations I Pg. 6 A. 1. x=1. x=. x=/ B. 7. x=/ 9. x=/ 11. x=-8 1. x=1 1. x=7 C. 17. x=- 19.x= 1. x=-1/.. x=- D.. x=/ 7. x=8/ 9. x= 1. x=0. x=. x=1 Linear Equations II Pg. 9 A. 1. x=.x=.x=- 7.x=/ 9.x=-7/ B. 11.x=1/ 1.x=1/ 1.x=-9/1 17.x=0/ 19.x=-/ 1.x=1.x=90/9.x=1 7.x=11/70 C. 9. x=1 1.x=0 D.. Identity. Inconsistent 7. Identity Linear Word Problems Pg.. A. 1. (x 10). (x +10). [x + (x 100)] = x x 9. x +8(x +) 11. (0 x) lbs of cashews, cost (00 x) 1. A gets $7, B gets $ 1. A gets $8, B gets $1, C gets $8 17. A gets $19.0, B gets $7.60, C gets $ $ gal.. $,000 at % and $6,000 at 6%. $7,00 at each 7. 8 lbs. of peanuts, lbs of cashews 9. lbs. of peanuts and 10 lbs. of cashews Literal Equations Pg. A. 1. b=a-d+e. b= (e-d) /. b= (e-d)/(+a) 7.y=/(x+6) 9. y=(-x)/6x 11.t=6w 1. t=6w+ 1.t=(aw+b)/ 17. t=6(w-a)+ 19.t=[c(w-d)+b]/a Inequalities Pg x <. x < 8. x 8 7. x 8 9. x < x < No such x 1. 7 x 17. x < x < 1. x. x 8/. / x 7. x 7/ / 7/ 8/ 100

101 The Rectangular Coordinate System Pg A(, 0), B(, 0), C(0, ), D(0, ), E(, ), F(, ), G(, ).. rd quadrant 7. Any ordered pairs with (+#, #) Linear Equations and Lines Pg. 1. y =. x =. y = / 7. x = 9. y = 8/ 11. y = / 1. x =

102 Systems of Linear Equations Pg. 7 A. 1. a.. c. B.. (0, ) 7. (, 1) C. 9. x 11. x 1. 9y D. 1. (1, ) 17. (, 0) 19. (, /) 1. (1, 1). ( 18, ).(, 1) 7.(, 7) E. 9. Dependent Systems of Linear Equations II Pg (1, ). (, 0). (, /) Linear Word Problems II Pg A gets $7, B gets $. A gets $86, B gets $. lb cashews, 8 lb peanuts 7. $11,0 at %, $0,0 10