SUNY ECC. ACCUPLACER Preparation Workshop. Algebra Skills

SUNY ECC ACCUPLACER Preparation Workshop Algebra Skills Gail A. Butler Ph.D.

Evaluating Algebraic Epressions Substitute the value (#) in place of the letter (variable). Follow order of operations!!! E) + 5; = E) n (n ) 5; n = - () () + 5 ( ) (( ) ) + 5 6 - + 5 (6)( 8 ) + 5-6 + 5 (6)(-9) + 5 - - + 5-7 E) p q : p = 5, q = - E) w 5w ; w = (5) ( ) () 5() + 5 9 6 0 + 6-6 + - E) ( ) (y ) ; = -, y = ( ) (() ) ( 5) (6 ) (5) 75 7 Practice Problems: Evaluate the given epression for the value(s) of the variable(s). Evaluate + for =. Evaluate y + 7 for = and y =. Evaluate ( + ) (y + 5) for = and y = Solutions:. -5. 5. -6 Combining Like Terms like terms: have eactly the same variable factors E. In the first eample above, and are like terms

To simplify an algebraic epression:. Remove parenthesis by distributing. Combine like terms (combine coefficients according to their sign; keep the variables and eponents the same.) E. (5y ) (6y ) 0y 6y Distribute 6y 9 Combine like terms Additional Eamples. Simplify the following algebraic epressions: a. + 8 The like terms are and ; and -8 b. ( 7) 6 Distribute. There are no like terms so we are done. c. 5 [7 0] Distribute the - to eliminate the [ ] 5 7 0 Combine like terms 7 5 d. 8 5y 9 + y - - y Practice Problems: Simplify the following epressions:. 6 y + 9y 0. 7(9 8) (y ). ( + 0) + ( + ) Solutions. 0 y 0. 6 y -0. + + 0

Solving Linear Equations The steps to solve a linear equation in one variable can be summarized as follows:. Simplify the algebraic epression on each side by distributing (if parenthesis are present) and combining like terms.. Move all of the terms containing variables to one side and all of the constant terms (numbers) to the other side. This is accomplished by adding or subtracting the same amount to each side of the equation.. Isolate the variable (usually accomplished by division or multiplication).. Check the solution in the original equation. Eamples: a. 5 = Add 5 to each side +5 +5 = 6 Divide by = To check, substitute into the original: () 5 = 6 5 = = b. 5 No grouping symbols are present, no like terms to combine on either side. Start with step : subtract from each side of the equation. 5 - - + 5 = -5-5 = 8 Subtract 5 from each side of the equation. To check: substitute = 8 into the original equation: 5 (8) 5 (8) + 5 = 6 + 9 = 9 The solution is correct because the same amount was obtained on both sides of the equation.

c. ( ) ( ) Begin by distributing and combining like terms on each side of the equation. + = - - Add to each side. + + + = - Subtract from each side. - - = - Divide by = - Check: ( ) ( ) ( ) ( ) ( ) (0) - + = - - = - Practice Problems: Solve the following equations for.. - = ( 8). ( + ) = ( 6). 5( ) = 6 + 0 9 Solutions.. 0 7. All real numbers Linear Equations with Fractions Equations are usually easier to solve if they do not contain fractions. To eliminate fractions from an equation, multiply both sides by the LCD (of all epressions). If you choose the correct LCD, no fractions will be present after you have multiplied. Eamples: a. n Multiply both sides (every term) by 5 (the LCD). 5 5 ( 5 ) + 5n = 5( ) + 5n = 5 Subtract from each side. - - 5n = -7 Divide both sides by 5. 5 5 5

n = 7 5 b. 7 Multiply both sides by (the LCD). 7 7 ( ) ( ) Note that needs to be distributed on each side. 7 7 7 7 8 7 7 8 7 9 Combine like terms 7 9 Subtract 7 from each side -7-7 7 = 9 Divide by 7 7 = 7 Practice Problems: Solve the following equations:. 5 =. + 8 = 8. 5 6 + 7 = Solutions. 5/8.. 7 6

Solving Inequalities A linear inequality in can be written as a + b < 0. The inequality sign might be any of the following: < less than > greater than less than or equal to greater than or equal to To solve a linear inequality, we can apply any of the procedures that we have already discussed for linear equations with the following eception: When an inequality is multiplied or divided by a negative number, the direction of the inequality sign is reversed. Why? Consider 5 < 0. Multiply (or divide) both sides by -. Is -5 < -0? The relationship between the numbers has changed and the correct relationship is -5 > -0. Eamples: Solve the following inequalities. Use interval notation to epress the solution set. Graph each solution set on a number line. a. + 5 < 7 Subtract 5 < Divide by < 6 b. 5 0 Divide by -5. Remember to reverse the direction of the inequality sign because you are dividing by a negative number!!! 6 c. ( ) 0 Distribute - 8 > + 0 Add -8 > 7 + 0 Subtract 0-8 > 7 Divide by 7 - > This should be rearranged so that is read first. This inequality is equivalent to: < - 7

d. 6( ) ( ) 7 8 6 6 7 8 7 0 7 8 0 8 Note that after subtracting 7 from each side, the variable has dropped out. Decide whether the remaining inequality is true or false. If true, the solution set would be all real numbers. If false, there is no solution set and we write Ø. (This is false so there is no solution.) e. 6 0 Subtract from each side. -6 < 0 Note that again, the variable has dropped out. This time, the resulting inequality is true. Therefore, the solution set is all real numbers. Practice Problems: Solve the following linear inequalities. + 0. (y + ) < ( y + 5). ( + ) 5. (8 + ) < ( 6 + ) Solutions:.. y 8 5. No solution. All real numbers Translating Words to Algebraic Epressions Key Words Addition: sum, added to, total, more than, greater than, increased by Subtraction: difference, subtracted from, less than, decreased by Multiplication: product, times, multiplied by, twice, of Division: quotient, divided by Remember: Switch order for less than and subtracted from Eamples 8 more than five more than twice a number + 8 5 + 7 subtracted from a number twice a number subtracted from ten 7 0 a number decreased by 6 the difference between and 7 6 7 8

the product of eight and a number four times the difference of a number and 8 ( ) 0% of a number twice a number increased by eight 0.0 + 8 the quotient of a number and 5 5 a number divided by three one eighth of a number si less than m m 6 8 three less than four times a number twice the difference of a number and four ( ) Epressions Involving Percents The cost, increased by 8.75% ta: The cost reduced by 0% : + 0.0875 0.0 Consecutive Integers consecutive integers :, +, + consecutive even integers, +, + consecutive odd integers, +, + Write each of the following as an algebraic equation:. Si less than a number is 6 =. Four times the difference of a number and 9 is 7. ( 9) = 7. Seven more than three times a number is two times the sum of the number and five. 7 + = ( + 5). The sum of a number and the number increased by 7 is 6. + ( + 7) = 6 9

5. Si less than three times a number is one-fourth the number. 6 = Practice Problems: Translate the following into algebraic epressions or equations. The product of a number and 8.. The difference of 6 and a number.. The sum of twice a number and 7. One sith of a number 5. Four less than a number 6. 60% of a number 7. Eight less than a number is fifteen 8. The sum of a number and the number decreased by 8 is 9. Nine less than si times a number is three times the difference of the number and seven Solutions. 8. 6. + 7. 6 5. 6. 0.60 7. 8 = 5 8. + ( - 8) = 9. 6 9 = ( 7) Solving Application Problems Problems that are presented to us in verbal form can usually be translated into a mathematical equation that can be solved to find the solution to the original problem. Follow the strategy below to help with this process:. Read the problem carefully! It is sometimes helpful to underline key words that indicate a certain mathematical operation.. Define a variable for one of the unknown quantities. Use the statement Let =.. If possible, write epressions for any other unknown quantities using the variable chosen in step.. Write an equation that represents the relationship between the unknown quantities. 5. Solve the equation. Use this solution to answer the question posed by the problem. 6. Check your solution according to the original wording of the problem. Note: If you check only in your equation and you have made an error in forming this equation, you may not notice this error. 0

Eamples: a. When seven times a number is decreased by, the result is. What is the number? Let = the number (Let = the unknown quantity.) 7 = (Write an equation.) + + (Solve) 7 = 7 7 = The number is. (Answer the question) Check: Seven times decreased by is ( = ) b. When 0% of a number is added to the number, the result is 5. What is the number? Let = the number (Let = the unknown quantity.) 0.0 + = 5 ***Remember, 0% of a number means 0.0 times the number.0 = 5.0.0 = 80 The number is 80. Check: 0% of 80 added to 80 = 5 0.0(80) + 80 = 5 7 + 80 = 5 5 = 5

c. One number eceeds another by. The sum of the numbers is 58. What are the numbers? Let = the first number + = second number ( eceeds means is more than: addition) Note that in this problem, there are unknown quantities, therefore a second epression is used to represent the second quantity. + ( + ) = 58 + = 58 - - = = 7 ( Sum indicates addition.) To find the second number, replace with 7 in the epression +. + = 7 + = The numbers are {7, } Check: 7 + = 58 d. In 00, the price of a sports car was approimately $80,500 with a depreciation of $765 per year. After how many years will the car s value be $6,975? Let = # of years until value is $,0 ( Depreciation means that the car is losing value each year. Be sure to multiply the loss each year (8705) by the number of years ()). 80500 765 = 0-80500 -80500-765 = -860-765 -765 = 5 After 5 years the car s value will be $0 Check: 80500 765(5) = 0 80500 860 = 0 0 = 0

e. After a 0% reduction, you purchase a dictionary for $5.0. What was the dictionary s price before the reduction? Let = price before reduction To calculate a percentage reduction, multiply the percentage by the original price, then subtract from the original price. 0.0 = 5.0 0.80 = 5.0 0.80 0.80 =.75 The original price was $.75. Check:.75(0.0) = 6.5.75-6.5 = $5.0 f. The selling price of a scientific calculator is $5. If the markup is 5% of the dealer s cost, what is the dealer s cost of the calculator? Let = dealer s cost To calculate a percentage markup, multiply the percentage by the original price, then add the original price. + 0.5 = 5.5 = 5.5.5 = The dealer s price is $ Check: (.5) = + = 5 g. The length of a rectangular pool is 6 meters less than twice the width. If the pool s perimeter is 6 meters, what are its dimensions? Let = width 6 = length ***Note less than requires us to subtract from

P = l + w 6 = ( 6) + 6 = + 6 = 6 + + 8 = 6 6 6 = The width is The length is () = 6 = 6 6 = 0 Check: () + (0) = 6 + 80 = 6 Practice problems: Solve the following problems:. When eight times a number is increased by, the result is 6. Find the number.. When 0% of a number is subtracted from the number, the result is 5. Find the number.. Two numbers differ by eight. The sum of the numbers is. Find the numbers.. After a 0% discount, the price of a new cell phone is $50. Find the price of the phone before the discount. 5. The length of a rectangular field is si feet greater than twice the width. If the perimeter is 8ft, find the length and width of the field. 6. A new pair of sneakers cost $75.5 after ta (rate of 8.75%) has been added. Find the price of the sneakers before ta. (Round your answer to the nearest cent.) Solutions.. 50. 7 and 5. $50 5. L = 78ft; w = 6 ft 6. $69.8 Rules of Eponents m n mn The Product Rule: b b b When multiplying eponential epressions with the same base, add the eponents.

Eamples: a. 5 6 b. 6 6 c. (remember: really has an eponent of that is added to the eponent -) d. ( 5 7 )(9 ) 99 e. 7 ( 5 y)( 6 y ) 0 y m b mn The Quotient Rule: b b 0 n b When dividing eponential epressions with the same base, subtract the eponents. Eamples: a. 8 0 7 5a b b. 0 7 c. 5a b 0 7 7a b The Zero- Eponent Rule: If b is any real number other than 0, b 0 Eamples: a. ( 9) 0 b. () 0 c. 0 (note: the eponent applies only to the, NOT to the ) d. 9 0 (note: the eponent applies only to the 9, not to the -.) The Negative Eponent Rule If b is any real number other than 0 and n is a natural number, then Eamples: b n n b a. 6 6 6 b. c. ( ) ( ) 9 5

m n mn The Power Rule: ( b ) b When an eponential epression is raised to a power, multiply the eponents. Eamples: a. 9 ( ) b. ( 55 ) 5 6 c. ( ) n n n Products to Powers: ( ab) a b When a product is raised to a power, raise each factor to that power. Eamples: a. 8 ( 6 ) 6 ( ) 6 b. 6 6 8 ( y ) ( ) ( ) ( y ) 7 y Quotients to Powers n a n a If b is a nonzero real number, then ( ) n b b When a quotient is raised to a power, raise the numerator and denominator to that power. Eamples: 6 ( 6) 6 a. ( ) b. y y y ( ) ( ) 8 6 Simplifying Eponential Epressions An epression is simplified when:. No parenthesis appear. No powers are raised to powers. Each base occurs only once. No negative eponents or 0 eponents appear. Eamples: a. 5a b 7 7a b 6 7 5a b 7 5a b (Divide the coefficients, subtract eponents.) b. (0 ) 6 (0 ) 000 third power in the denominator.) (Eliminate the negative eponent, raise each factor to the 6

Practice Problems Simplify the following:. 8 y 8 7 y. ( ) (y ). (5 y )( 7 y 0 ). (5y ) 5. ( 5 y ) 0 Solutions.. y 5 7 y 7. 5 y. 5y 6 5. Multiplying Polynomials (Monomial)( Monomial): Multiply coefficients, add eponents of like bases. Eamples: a. ( 5 8 6 )( ) 8 b. 8 0 8 ( y )( 6y ) y (Binomial)(Binomial) We can use the acronym FOIL to describe how to multiply two binomials: E. ( + 6)( 5) F stands for the first terms in each binomial: O stands for the outer terms: 5 5 I stands for the inner terms: 6 6 L stands for the last terms in each binomial: 6 5 0 Combine like terms and write your answer in standard form: 5 6 0 0 Eamples: a. ( 5)(7 + ) ( )(7) ()() ( 5)(7) ( 5)() 6 5 0 9 0 Practice combining the outer and inner terms (if like) mentally and just writing the final answer. It will be important to be able to perform this multiplication quickly as you progress through mathematics courses. b. (7 )( 5) 5 6 0 0 7

c. (7 5)( ) 5 7 5 0 Note: there are no like terms to combine here. Practice Problems Multiply the following:. ( 6)( + 8). ( 9)( + 5). (y 5)(y + 7). ( + 7)(y 5) Solutions:. + 8. 6 7 5. 6y + y 5. 8y 0 + 8y 5 Factoring Polynomials Factoring is the process of breaking a polynomial into a product. It is the reverse of multiplying. Common Factors Step of factoring is always to remove the GCF (greatest common factor). Remove means divide each term by the GCF. E. 6 The GCF of 6 and - is 8. Write the GCF outside a set of parenthesis. Mentally divide each term by 8. Write each resulting quotient inside the parenthesis. Ans: 8( ) Note: This answer can be checked by distributing. When the is distributed, the result should match the original problem. If it does not, the factorization is not correct. E. 6 8 The GCF is 6. Note that when variables are involved, use the smallest eponent of each common variable as part of the GCF. 6 ( ) (Remember to subtract eponents when dividing like bases). Practice Problems Factor each of the following:.. 8y y. 0 + 8

Solutions. ( ). y(y ). ( 5 + ) Factoring Trinomials of the Form a b c with a = Remember, factoring is the reverse of multiplying. Eamine these multiplication problems: a.( )( 5) b.( + 6)( + ) c. ( + )( ) d. ( + 5)( ) 8 5 7 6 0 Our goal is to get from these answers back to the factored form. Use the Trial and Error method, incorporating some guidelines. E. 7 6 (a =, b = 7, c = 6) Form of answer: ( + )( + ) We must fill in the blanks with factors whose product is c Factors of 6 (c): 6 Which pair adds up to 7 (b): 6 ( + 6)( + ) Check by multiplying (FOIL) Eamples: a. 8 5 (a =, b = 8, c = 5) Form of answer: ( + )( + ) We must fill in the blanks with factors whose product is c Factors of 5 (c): 5 5 Which pair adds up to 8 (b): 5 ( + 5)( + ) Check by multiplying (FOIL) Note that ( + )( + 5) is also a correct answer. The order of the factors in the answer does not matter. b. 5 Factors of 5 that add up to - are -9 and -5 ( 9)( 5) 9

c. 5 Factors of -5 that add up to - are -5 and + ( 5)( + ) d. 0 Factors of -0 that add up to + are +5 and - ( + 5)( ) Note: All of these answers can be checked by FOILing!!! ****Notice that when the sign of c (the constant) is +, the signs in each factor of your answer will be the same; the sign of b (the middle term) will tell you if both are or +. ****Notice that when the sign of c (the constant) is -, the signs in each factor of your answer will be different; the sign of b (the middle term) will tell you the sign of the larger number. Practice Problems. + + 8. y + y 0. y 5y + 56. 7 0 5. y y 8 Solutions:. ( + )( + 9). (y + 5) (y ). (y 8)(y 7). Prime (cannot be factored) 5.(y )(y + ) Factoring Trinomials of the form a b c with a Use trial and error but there will be more possibilities to consider. E. 5 The only way to break up Our possibilities are: is. The only way to break up is. ( + )( + ) OR ( + )( + ) A quick check of the inners and outers (from FOIL) will determine the correct factorization. They must produce 5 when combined. ( + )( + ) OR ( + )( + ) inners:, outers: inners:, outers: 6 combined: 5 combined: 7 The first factorization is correct!! 0

Eamples: a. 6 7 The possible factors of 6 are 6 and. The possible factors of are, 6 and (or each pair with both negative factors). The sign of b (-7) is negative indicating that the factors of will both be negative. Trial : ( )( ) outers: - inners: -6 combined: -8 (incorrect) Trial : ( )( ) outers: -9. inners: -8 combined: -7 (correct) b. 9 5 The possible factors of and. 9 are 9 and. The possible factors of - are,, Trial : ( )( + ) outers: inners: - combined: -9 (incorrect) Trial : (9 + )( ) outers -6 inners: combined: -5 (incorrect) Note: After checking the outers, you might quickly decide that it appears too large to continue and go on to the net trial. Trial : (9 - )( + ) outers: 9 inners: - combined: 5 (correct) Practice Problems:. 5. 5y y +. + 0 + 7 Solutions:. ( + 5)(( ). (5y )(y 6). ( + )( + ) Factoring the Difference of Two Squares When factoring an epression containing terms separated by a - sign, each of which is a perfect square, the factorization will always be of the following form: A B ( A B)( A B) Note: the factors in your answer can be in either order. It doesn t matter if the + is first or the - is first.

Eamples: a. A variable epression containing an even eponent is a perfect square, is also a perfect square. ( + )( ) b. 6 y 9 = (6 + 7y)(6 7y) Practice Problems. 5. y 8. 00 Solutions. ( 5)( + 5). (y 9)(y + 9). (0 - )(0 + ) Repeated Factorizations In order to completely factor an epression, it must be broken down as far as possible. This often requires more than one factoring step. Always remove the GCF first if possible. Eamples: a. 8 First, recognize the epression as the difference of squares (9 )(9 ) Note that the second factor is again the difference of squares and should be factored again. (9 )( )( ) b. 50 0 First, remove (divide by) the GCF of (5 5 6) Continue factoring by trial and error (5 + )(5 ) Practice Problems. 8 9y. 8. + 6 Solutions:. (9( + y)( y). ( + )( ). ( + )( )

Rational Epressions Simplifying a Rational Epression To simplify a rational epression:. Factor the numerator and denominator completely. Divide both the numerator and denominator by any common factors.. State any numbers that are ecluded from the domain of the original epression. If the epression is simplified, the numerator and denominator will not contain any common factors other than. Eamples: Simplify. 8 ( ) a. Factor the top and bottom ( )( ) = The top and bottom were divided by the common factor: (-). The restrictions on the domain were stated. y y 5 ( y 5)( y ) y 5 b. y, (Note: the restrictions include ALL y 5y ( y )( y ) y values that make the original epression undefined, not only those that make the simplified version undefined. Practice Problems: Simplify the following epressions:. + 8 Solutions:. y 7y+0 y y 0. +0. +. y y+. 5

Multiplication and Division of Rational Epressions Multiplying Rational Epressions. Factor all numerators and denominators completely.. Divide numerators and denominators by common factors (you can divide one factor anywhere on the top by the identical factor anywhere on the bottom.). Multiply the remaining factors in the numerator; multiply the remaining factors in the denominator. Eamples: a. Factor ( )( ) ( ) ( )( ) Divide the common factors ( ), ( ) and ( + ), b. 6 9 7 ( )( ) ( )( 9) ( 9) Dividing Rational Epressions To divide rational epressions, multiply the first epression by the reciprocal of the second epression (as we do with rational numbers). Remember, the reciprocal of an epression is formed by inverting (flipping it upside down). Follow the steps for multiplication of rational epressions. Eclude from the domain, any value(s) that makes the original epressions undefined, as well as any value(s) that makes the inverted epression undefined.

Eamples: a. 8 8 Keep the first epression, multiply by the reciprocal of the second. Factor ( )( ) ( ) Divide the common factors ( ), b. 5 6 5 6 Keep the first epression, multiply by the reciprocal of the second. Factor ( ) ( )( ) ( )( ) ( )( ) ( ) ( ) Divide the common factors Multiply numerators and denominators. ( ) ( )( ),,,, Practice Problems: Perform the indicated operation and simplify.. +7+ 6. (+) 5 0 9. 6 5 +0 +6. 5 y 8 z y z Solutions:. + 6. (+) ( ). + 5. 5 y 7 8 5

6 Addition and Subtraction of Rational Epressions Same Denominator Add or subtract the numerators, the denominator remains the same; simplify. Eamples: a. 6 6 Add the numerators, denominator remains the same 6 6 Factor to simplify. ) )( ( ) )( ( Divide the like factors., b. 6 6 Subtract numerators (remember to distribute the - sign). 6 6 6 ) ( Factor to simplify ) ( Different Denominators The epressions must be rewritten as equivalent epressions containing the least common denominator (LCD). To find the LCD:. Factor each denominator completely.. List the factors of the first denominator.. Add to the list, any factors of the second denominator that do not appear in the list (there is no need to repeat any common factors).. The product of all the factors in the list is the LCD of the epressions.

Eamples: a. Find the LCD of and 0 ( 5)( ) 6 ( )( ) Factor each denominator completely. Factors of first denominator: ( + 5) ( ). From the second denominator, add ( + ) to the list. So the LCD is ( + 5)( )( + ). Note: these factors can appear in any order but all three must appear in the LCD. 5 b. Find the LCD of and 5 5 Factor each denominator completely. 5 5 (5 )(5 ) Factors of first denominator: (5 + ). From the second denominator, add (5 ) to the list. (There is no need to repeat the identical factor.) So the LCD is (5 + )(5 ). Note: these factors can appear in any order. To add or subtract rational epressions that have different denominators:. Find the LCD of the rational epressions.. Rewrite each epression as an equivalent one using the LCD. (Multiply the numerator and denominator by any factor needed to change the denominator into the LCD.). Add or subtract the numerators; keep the denominator the same.. Simplify if possible. 7

Eamples: 8 a. LCD is ( )( + ). To find the equivalent epressions, multiply the first fraction by ( + ) and the second fraction by ( ). Remember, multiply BOTH the numerator and denominator. 8( ) ( ) ( )( ) ( )( ) Add the numerators (remember to distribute!) 8( ) ( ) ( )( ) simplified. 8 0 0 ( )( ) ( )( ) Factor to see if the epression can be 0( ) ( )( ), There are no common factors to divide. b. 7 6 Factor to find the LCD. ( 6)( ) ( 6)( ) LCD: ( 6)( + )( ) To find the equivalent epressions, multiply the first fraction by ( ) and the second fraction by ( + ). Remember, multiply BOTH the numerator and denominator. ( ) ( ) ( 6)( )( ) ( 6)( )( ) Subtract the numerators (remember to distribute!) ( ) ( ) ( 6)( )( ) ( 6)( )( ) 5 ( 6)( )( ) 5,, Practice Problems: Perform the indicated operation and simplify.. 5 + +8 +8 Solutions:. 7 + +. + 6 + 8+5. 5 8 + 5. + 5 7. 5. 9+8 (+). 8 ( )( 5). +8 ( )(+) 5. (+)( 7) 8

Solving Rational Equations A rational equation contains one or more rational epressions (remember, these contain variables in the denominator). Follow the same procedure for solving as above with linear equations containing fractions. Sometimes, one or more denominators will need to be factored in order to determine the LCD. Remember to avoid any values that make any rational epression involved undefined (values that make the denominator = 0). Eample: 5 8 LCD: 8 (all denominators divide evenly into 8). Multiply 9 8 both sides by 8. 5 8 8( ) 8( ) Note that 8 needs to be distributed on each side. 9 8 5 8 8 8 8 8 9 8 5 8 9 6 5 6 6 Add 6 to each side. + 6 +6 5 = 7 6 Add 6 to each side +6 +6 5 = 7 Divide by 7 7 7 = Check: 5 8 9 8 5 5 6 5 8 9 8 9 6 8 8 8 9 8 8 8 8 8 9

Practice Problems: Solve the following equations.. 0 + = 0 Solutions:. 6 + = 5. + = 7 6. 0, -5. 0. Comple Fractions (Comple Rational Epressions) Simplifying To simplify a comple rational epression:. Find the LCD of all the rational epressions in the numerator and denominator.. Multiply each term of the epression by the LCD (this should eliminate all of the rational epressions).. Simplify the resulting epression by factoring (if possible) and dividing out common factors. Eamples: a. The LCD is. Multiply each term by. 6 Factor ( ) b. y y The LCD is y. Multiply each term by y. y y y y y y y 0, y 0 (No factoring is possible.) 0

Practice Problems: Simplify the following rational epressions:. + +. + 5. y + y Solutions:. +. + (5 ). y y+ Solving Systems of Linear Equations Substitution Method Steps:. Solve one of the equations (either one) for one variable (either one).. Substitute the epression obtained in Step into the other equation. (only one variable should now be present.). Solve this resulting equation.. Re-substitute the answer obtained in Step into the either original equation. Solve for the other variable. 5. Check the solution in BOTH original equations. Eamples: Solve the following systems of equations by substitution. a. y = - () (The equations are numbered so that they y = + 7 () can be easily referred to.) Equation () is already solved for y so step can be omitted. The epression + 7 will be substituted into Equ. () in place of y: y = - ( + 7) = - Distribute and solve for. 6 = - - = - - = 8 = - Re-substitute into equ () (or equ () I just chose equ ()) to find y: y = + 7 y = (-) + 7 y = - + 7 y = Solution (-, )

Check: y = - y = + 7 (-) () = - = (-) + 7-9 = - = - + 7 - = - true = true Note: When both lines are graphed, the point of intersection is (-, ) as can be seen on the graph below: b. + 5y = () - + 6y = 8 () It looks like it will be easiest to solve equ. () for (the coefficient is - which should not create any fractions when solving.) - + 6y = 8 - = 8 6y Divide all terms by - = -8 + 6y = 6y -8 Substitute into equ. (): + 5y = (6y 8) + 5y = y 6 + 5y = 7y 6 = 7y = 7 y =

Re-substitute into equ. () (or equ (). Note: As we have already isolated in our first step, we could also re-substitute into that epression instead ( = 6y 8) as long as we are sure that no mistakes have been made in isolating. - + 6y = 8 - + 6() = 8 - + 6 = 8 - = = - Solution: (-, ) Check: + 5y = - + 6y = 8 (-) + 5() = -(-) + 6() = 8 - + 5 = + 6 = 8 = true 8 = 8 true c. y () y 7 () Both equations have one variable already solved for. Let s substitute equ. () into equ. (): y 7 Multiply all terms by to clear fractions. + 8 = - + 8 Solve for 5 = -0 = - Re-substitute into equ. () to find y. y 7 y ( ) 7 y = - + 7 y = Solution (-, ) Check: y y 7 ( ) ( ) 7 7 = true = true

Practice Problems Solve the following systems by substitution:. y =. y = 7. y = 5y = + y = 6 0 + ½ y = Solutions:. (, ). (, ). (/8, -/) Addition Method Some tetbooks refer to this method as solving by elimination, meaning that one variable is eliminated (usually by addition). In this method, the like terms in each equation are lined up, added (so that one variable is eliminated) and the resulting equation is solved. This answer is then substituted into either original equation to find the other value. An eample of the simplest type is provided below with more detailed steps to follow: E. Solve by addition: + y = 6 y = - Add the equations together (notice that the ys drops out). + y = 6 y = - = = Solve for Substitute into either original to find y: + y = 6 + y = 6 y = - The solution is (, -) It can be verified that this solution is correct when it is substituted into BOTH of the original equations. Sometimes adding the original equations will not result in one of the variables being eliminated. We can adjust one (or both) equations to force this to happen. To do this, multiply one (or both) by nonzero constants so that the coefficients of one of the variables are opposites. Then, when the equations are added, that variable will be eliminated.

E. + 5y = () - + 6y = 8 () If equ () is multiplied by (every term), the system becomes: + 5y = - + y = 6 Add the resulting equations: 7y = 7 y = Substitute into either original equation (I ll use equ ) + 5() = + 5 = = - = - Solution: (-, ) Check this answer in BOTH original equations to be sure that it is correct. Summary of Steps:. Rewrite (if necessary) each equation in the form A + By = C (this is so like terms, and the = signs are lined up).. If necessary, multiply one or both equations by a nonzero constant (so that the coefficients of one of the variables are opposites).. Add the resulting equations.. Solve for the remaining variable. 5. Substitute into either of the original equations to find the other variable. 6. Check the solutions in BOTH of the original equations. Eamples: Solve the following equations using the addition method. a. 7y = () 6 + 5y = 7 () In this case, we can multiply equ () by - to make the coefficients of opposites. To indicate this in your work, write the following: -( 7y = ) 6 + 5y = 7-6 + y = -6 6 + 5y = 7 Add the equations 9y = -9 y = - Substitute into equ () (or equ ()) 5

7(-) = equ () + 7 = = 6 = Solution: (, -) Check: 7y = 6 + 5y = 7 () 7(-) = 6() + 5(-) = 7 6 + 7 = - 5 = 7 = 7 = 7 a. + y = -6 5 -y = 0 (Multiply the top equation by -5 and the bottom equation by ) -0 5y = 80 0 0 y = 60 (Add) -5y = 0 (Divide by -5) y = - Substitute into either original equation (I used the top one): + () = -6 = -6 = - = - The solution is (-, -). Check this in both original equations. Practice Problems: Solve the following systems of equations by the Addition method. y = -. y =. + = -5y -+6y = -6 + 5y = 6 8 = 7 Solutions:. (-6, -). (, ). (8, -5) 6

Linear Systems with No Solution or Infinitely Many Solutions It is possible that a linear system has no solution (no ordered pair satisfies BOTH equations at the same time) or infinitely many solutions (there is an endless list of ordered pairs that satisfy both equations). If the equations were graphed on the same set of aes, a system with no solution would be two parallel lines. A system with infinitely many solutions would be the same identical line and any point on that line is a solution. E. Solve the following system by addition: 6 + y = 7 y = Line up in A + By = C form: 6 + y = 7 () + y = () Multiply equ () by - 6 + y = 7-6 y = - Add 0 + 0 = 0 = Both variables were eliminated and the resulting equation is False. Therefore, this system has no solution. We can use the symbol for empty set to indicate this: Ø E. Solve the following system by addition: y = () y = () Multiply equ () by-: y = - + y = - Add 0 + 0 = 0 Both variables were eliminated and the resulting equation is True. Therefore, this system has an infinite number of solutions. Set builder notation (using either original equation) is often used to indicate all of the solutions: {(, y) y = } OR {(, y) y = } These are read the set of all ordered pairs (, y) such that y = OR the set of all ordered pairs (, y) such that y = meaning that the equation in the solution must be satisfied by the ordered pair in order to be a solution. Practice Problems Solve the following systems by either method. y = 7. + y = + 8y = - - y = 5 7

Solutions:. Infinite number of solutions. No solution Radicals Product Rule for Square Roots (a, b are nonnegative real numbers) ab a b and a b ab The square root of a product is the product of the square roots. Eamples: a. 6 9 6 b. 5 00 0 We will use this property to simplify square roots. Simplifying Square Roots To be simplified, the radicand must have no perfect square factors other than. Break the radicand into the product of a perfect square and another number, simplify by removing the square root of the perfect square. Eamples: a. 7 9 9 is a perfect square that is a factor of 7. Its square root is which is written on the outside of the radical sign. b. 5 55 5 5 c. 7 6 6 Note: the last eample could have also been simplified by the following series of steps: 7 98 8 6 If you do not use the largest perfect square that is a factor at first, just keep simplifying until there are no more perfect square factors. If you remove another square root, multiply it by the number that is already outside the radical sign (when the was removed, it s value was multiplied by the that was already outside). Note: Algebraic epressions with even eponents are perfect squares. 8

Eamples: a. b. (because 8 (because ) 8 ) Use this fact (in conjunction with the rules already learned) to simplify the following epressions: a. 0 8 80 6 5 5 First, the radicands are multiplied; net the radicand is broken into factors using one that is a perfect square, simplify. b. 6 8 9 Practice Problems: Simplify the following radical epressions:. 90 6 y 8. 08 y. 0y 8 5 Solutions:. 8 y 0. 6y. y 0 9