CAHSEE on Target UC Davis, School and University Partnerships


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3 UC Davis, School and University Partnerships CAHSEE on Target Mathematics Curriculum Published by The University of California, Davis, School/University Partnerships Program 006 Director Sarah R. Martinez, School/University Partnerships, UC Davis Developed and Written by Syma Solovitch, School/University Partnerships, UC Davis Editor Nadia Samii, UC Davis Nutrition Graduate Reviewers Faith Paul, School/University Partnerships, UC Davis Linda Whent, School/University Partnerships, UC Davis The CAHSEE on Target curriculum was made possible by funding and support from the California Academic Partnership Program, GEAR UP, and the University of California Office of the President. We also gratefully acknowledge the contributions of teachers and administrators at Sacramento High School and Woodland High School who piloted the CAHSEE on Target curriculum. Copyright The Regents of the University of California, Davis campus, All Rights Reserved. Pages intended to be reproduced for students activities may be duplicated for classroom use. All other text may not be reproduced in any form without the express written permission of the copyright holder. For further information, please visit the School/University Partnerships Web site at:
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5 Introduction to the CAHSEE The CAHSEE stands for the California High School Exit Exam. The mathematics section of the CAHSEE consists of 80 multiplechoice questions that cover 5 standards across 6 strands. These strands include the following: Number Sense (14 Questions) Statistics, Data Analysis & Probability (1 Questions) Algebra & Functions (17 Questions) Measurement & Geometry (17 Questions) Mathematical Reasoning (8 Questions) Algebra 1 (1 Questions) What is CAHSEE on Target? CAHSEE on Target is a tutoring course specifically designed for the California High School Exit Exam (CAHSEE). The goal of the program is to pinpoint each student s areas of weakness and to then address those weaknesses through classroom and small group instruction, concentrated review, computer tutorials and challenging games. Each student will receive a separate workbook for each strand and will use these workbooks during their tutoring sessions. These workbooks will present and explain each concept covered on the CAHSEE, and introduce new or alternative approaches to solving math problems. What is Algebra? Algebra is a branch of mathematics that substitutes letters (called variables) for unknown numbers. An algebraic equation equates two expressions. It can be thought of as a scale that must remain in balance. Whatever is done to one side of the equation (e.g., addition, subtraction, multiplication, division) must be done to the other side. In this workbook, each unit will introduce and teach a new concept in algebra. 1
6 Unit 1: Foundational Concepts for Algebra.0 On the CAHSEE, you will be asked to solve linear equations and inequalities in one variable. The key to solving any equation or inequality is to get the unknown variable alone on one side of the equation. Example: x + 5 = 11 x = 6 x = Variable Isolated Equations must balance at all times. Therefore, whatever you do to one side of the equation, you must do to the other side as well. Example: x + 5 = 11 x = 115 x = 6 x 6 = x = Subtract 5 from both sides. Divide both sides by. x is isolated on one side of the equation. There are three important concepts that come into play when solving algebra equations: Adding opposites to get a sum of 0 Multiplying reciprocals to get a product of 1 Using the distributive property We will look at each of these individually and examine how they are used in algebra.
7 A. Taking the Opposite.0 Rule: To take the opposite of the number, simply change the sign. The number line below shows the relationship between opposites. The opposites of the positive integers are their corresponding negative integers; the opposites of the negative integers are their corresponding positive integers. Example: The opposite of +7 is  7. On Your Own: Find the opposite of each number. Number Its Opposite Adding Opposites Rule: The sum of any number and its opposite is always zero. Example: (100) = 0
8 .0 On Your Own: Fill in the blanks to make the following steps true: = 0. If y = 54, then y = 54. If x = 0, then x = If x = 1 1, then x = = = ,40 + (1,40) = 0 8. If x = 0, what is the value of x? If x + = 0, what is the value x? 10. If x + 95 = 0, what is the value of x? 95 4
9 B. Using the Concept of Opposites in Algebra.0 The concept of opposites is essential when solving algebraic equations. As we said at the beginning of this unit, the key to solving any equation is to get the unknown variable (for example, x) all by itself. Adding opposites (which gives us the sum of 0) is key to isolating an unknown variable. Let's look at the following example: Example: x + 5 = 7 In the above problem, we need to find the value of x, which is our unknown variable. This means that we need to get rid of the "5" on the left side of the equation so that the x stands alone. Since the sum of any number and its opposite is always zero, we can get rid of the "5" by adding its opposite. What is the opposite of 5? 5 Because equations must balance at all times, we need to add 5 to the other side of the equation as well: Let's look another problem: Example: x  = 9 To isolate x on the left side of this new equation, we must take the opposite of . The opposite of  is + (Remember: To take the opposite of any number, just change the sign.) So we must add + to both sides of the equation:  + = 0 x  + () = 7 + x = 10 5
10 On Your Own: Use the concept of opposites to solve for x. 1. x + = 1.0 x +  = 1  x = x = x = 11 + x = 14. x 11 = 7 x = x = 8 4. x  7 = 15 x = x = x = x = 4 + x = 1 6
11 C. Reciprocals.0 Definition: The reciprocal of a number is its multiplicative inverse. When we multiply a number by its reciprocal, we get a product of 1. Reciprocal means the flipside, or inverse. Example: The reciprocal of 5 4 is 4 5. Note: Zero does not have a reciprocal. On Your Own: 1. The reciprocal of 7 is 7. The reciprocal of 9 is 9. The reciprocal of 5 is 5 4. The reciprocal of 8 is 8 5. The reciprocal of 8 7 is The reciprocal of is 7. The reciprocal of 9 4 is 4 9 7
12 i. Finding the Reciprocal of a Whole Number.0 Example: The reciprocal of 6 is 6 1 because any whole number can be written as a fraction with a 1 in its denominator: (6 = 16 ); we just flip it to get the reciprocal! The reciprocal of a whole number will always have a numerator of 1! On Your Own: = 1 The reciprocal of 10 is = 1 The reciprocal of is = 5 The reciprocal of 5 is The reciprocal of 16 is The reciprocal of 18 is The reciprocal of 1 is The reciprocal of 14 is The reciprocal of  is 8
13 ii. Finding the Reciprocal of a Fraction.0 Rule: To find the reciprocal of a fraction, flip the numerator and the denominator. Example: The reciprocal of is Example: The reciprocal of 51 is 5 +1 = 5 On Your Own: Find the reciprocal of each number: , x  x² x 1_ x² iii. Multiplying Reciprocals Rule: Whenever we multiply two numbers that are reciprocals of one another, the product equals Example: 1 and are reciprocals because = = Example: 5 1 and 5 are reciprocals because = 1. On Your Own: Solve each equation. 19 = = 1 57 = = 1 ² 1 = 1 1 y = 1 y 9
14 D. Using the Concept of Reciprocals in Algebra.0 The concept of finding reciprocals is essential in algebra. In order to isolate an unknown variable from its coefficient (the numeric factor of the term), we need to multiply that coefficient by its multiplicative inverse (or reciprocal). Example: To isolate x, we need to multiply the coefficient by its multiplicative inverse (or reciprocal), which is 1. Since the equation needs to remain balanced at all times, we also need to multiply the term on the right side of the equation by the same value: ½ = 1 x 1 = x = 10 x = 5 Of course, we are really just dividing both sides by, but it is the concept of reciprocals that is at work here. 10
15 Let's look at another example:.0 Example: x = 9 Here, the coefficient is 1 and its reciprocal is 1, or. Therefore, to isolate x and keep the equation in balance, we must multiply both sides by : x = 9 ( x ) = (9) We multiply both sides by ( 1 x 1 )(x) = 18 Notice that can be written as x 1x = 18 x = 18 Now let's look at an example with a negative number: Example: x = 9 1 Here, the coefficient is and its reciprocal is . Therefore, to isolate x and keep the equation in balance, we must multiply both sides by : x = 9 ( x ) =  (9) x =
16 On Your Own: Use the concept of reciprocals to solve for x x = 1 x 1 = x = 7. 4x = 16 4x 16 = 4 4 x = x = 7 (5)( 5 x ) = (5)(7) x = 5 4. x = 8 x = x = 16 x = 16 1
17 E. Distributive Property.0 Example: (4 + 5) = ( 4) + ( 5) = = 7 According to the distributive property, multiplication may be distributed over addition. In other words, when multiplying a number or variable by two or more terms that are added or subtracted (and closed off in parentheses), the multiplication is distributed to each term. We then add the products. Note: Following the order of operations, you would get the same result. Solving what is in the parentheses first, you would get the following: Example: (4 + 5) = 9 = 7 On Your Own: Use the distributive property to find the value of each expression. 1. 4(5 + 9) = = 56. (91) = 7  = 4. (6  ) = 16 = (  5) = = ( + ) =  = 1 1
18 F. Using the Distributive Law in Algebra.0 The distributive law is used to simplify algebraic expressions like the following: Example: (x  4) In the above expression, the number outside of the parentheses () must be multiplied by each term within the parentheses: x and 4. (x  4) = ( x) + ( 4) = 6x + ( 1) = 6x  1 On Your Own: Use the distributive law to simplify each of the expressions. 1. (x + 5) = 4x (x  5) = 4x (x ) = 4x (x 4) = 6x (x + 4) = 4x (x + 14) = x  14 Note: The distributive law is also used to solve equations. This will be taught later. Note to Tutors: Concepts and skills from Unit 1 will also appear in the Unit Quiz. 14
19 Unit Quiz: 1. Solve for x: x + 5 = 4 x = 19. Solve for x: x = 54 x = 18. Simplify the expression (x + ) x Solve for x: x  = 4 x = 1 5. Simplify the expression (x  6) x Solve for x: (1x) = 6 x = Simplify the expression 4(4x + 5) 16x Solve for x: 5x = 5 x = 7 15
20 Unit : Solving Linear Equations 4.0 & 5.0 Now we are ready to solve a simple equation: Example: x + = 15 To get x all by itself, we must remove everything else on that side of the equation. x + = 15 x +  = 15  We apply rule of opposites x = 1 We combine like terms x 1 = x = 6 We apply rule of reciprocals Note: Unlike the order of operations, when isolating an unknown variable, we begin with addition or subtraction, and then move on to multiplication or division. On Your Own Example: 4 x  = 6 x  = 6 4 x  + = Apply rule of opposites x (4) = (4)(8) Apply rule of reciprocals 4 x = 16
21 Practice: Solve for x. 4.0 & x = 9 6x = 1 x =. x + 5 = 7 x = x = 4. x  4 = x = 6 x = 18 17
22 Isolating the Unknown Variable 4.0 & 5.0 As mentioned above, whenever we solve an algebra problem, our goal is to get the unknown variable, such as x, all by itself. (Note: We are solving for the unknown variable.) In problems in which the unknown variable appears on both sides of the equation, we need to move one of them to the other side. In other words, we need to get all the x's on one side and combine them. Look at the example below: x + 1 = 5x + 10 x = 5x We apply rule of opposites on numbers x = 5x + 9 We combine like terms x  5x = 5x  5x + 9 We apply rule of opposites on variable x = 9 We combine like terms x = 9 We multiply by reciprocal x =  On Your Own: Solve for x. 1. x + = x + 5 x  x + = x x + 5 x + = 5 x +  = 5  x = Apply rule of opposites Combine like terms Apply rule of opposites Combine like terms x = 18
23 . x = 4x & 5.0 x + 4x  = 4x + 4x + 7 5x  = 7 5x  + = 7 + 5x = 10 5x 10 = 5 5 Apply rule of opposites Combine like terms Apply rule of opposites Combine like terms Divide both sides by 5. x =. x + 8 = 4x x + 4x = 108 x = x = 1 x = x = x  x = 6 x = x = 19
24 Simplifying Linear Equations 4.0 & 5.0 Some problems will be slightly more complex than the ones we saw above. For these types of problems, we need to simplify first. Example: (x  5) + 4(x  ) = 1 Get rid of the parentheses. Use the distributive law: (x  5) + 4(x  ) = 1 6x x  8 = 1 Combine like terms: 6x x  8 = 1 6x + 4x = 1 10x  = 1 Now we have a simple equation to solve: 10x  = 1 10x  + = 1 + We add to both sides 10x = 5 10x 5 = We multiply both sides by (Divide by 10) 10 5 x = 10 This is an improper fraction: numerator > denominator Change to a mixed number and reduce: 1 0
25 On Your Own 4.0 & 5.0 Example: (x + )  (x + 4) = 15 Use distributive law to get rid of the parentheses. x x  1 = 15 Combine like terms: x 6 = 15 Solve: x 6 = 15 x = 1 x = 1 1
26 4.0 & 5.0 Practice: Simplify the following problems, and then solve. 1. 5(x  )  (x  4) = 5 5x x + 1 = 5 x  = 5 x = 8 x = 4. (x + 4)  (x + 6) = 1 x x  18 = 1 x 10 = 1 x = 11 x = (x + ) + (x 4) = 0 6x x 1 = 0 1x + 6 = 0 1x = 4 x =
27 Cross Multiplying 4.0 & 5.0 Some problems will be made up of fractions on either side of the equation. Before we can solve these types of equations, we need to get rid of the fractions. We do this by cross multiplying. Cross multiplying means multiplying diagonally across the equation sign. The term is shorthand for expressing the mathematical law governing proportions, according to which, "the product of the means is equal to the product of the extremes." We can cross multiply to find an unknown variable in a proportion problem: Example: Solve for x. 9 = 8 x x = 7 x = 4 When we cross multiply, we get x Now divide both sides by to solve for x.
28 Let's look at the following example: 4.0 & 5.0 x + 1 = x 5 Multiplying diagonally across the equation sign, we get the following: x + 1= x 5 5(x + 1) = (x) 5(x + 1) = (x) 5x + 5 = 9x Here, we multiplied to get rid of the parentheses. Now we have a regular equation to solve. When solving any equation, the goal is to get the variable, x, all by itself: 5x + 5 = 9x 5 = 4x Subtract 5x from both sides 5 = x Divide both sides by 4 4 Note: If you end up with an improper fraction (numerator > denominator), you need to change it to a mixed fraction: 5 1 = Note: On the CAHSEE, the answer might appear as a mixed fraction or an improper fraction. 4
29 On Your Own 4.0 & 5.0 x x + Example: Simplify the following equation: = 1 Cross multiply: x (1) = (x + ) Remove the parentheses: 4x = x + 9 Isolate the x: 4x = x + 9 1x = 9 x = 7 5
30 Practice: 4.0 & 5.0 x x = x() = (x + 5) x = 6x x = 15 x = 5. x + 1 = 4 x 4(x + 1) = x 4x + 4 = x x = 4 x = . x x 1 = 4 (x) = 4(x  1) 6x = 4x  4 x = 4 x =  6
31 Equations Involving Squares and Roots.0 & 4.0 On the CAHSEE, you may be asked to solve algebra problems that involve squares and roots. A square is a number raised to the second power. When we square a number, we multiply it by itself: Example: 5² = 5 5 = 5 The square root of a number is one of its two equal factors: Example: 5 = 5 because 5 is one of the two equal factors of 5. Note: 5 is also 5 because 55 is 5. However, on the CAHSEE, you will be asked to give only the positive root. On the CAHSEE, you might get a problem like the following: Solve for x: x² = 81 In order to isolate x, we must undo the squaring of x: This means perform the inverse operation. The inverse of squaring is taking the square root: x² = 81 x = 81 Take the square root for both sides. x = 9 7
32 Here's a slightly more difficult problem:.0 & 4.0 Solve for x: x 49 =  For problems involving roots, you should simplify before solving: x =  49 x =  7 Note: We took the square root of 49 ( 49 = 7) Now we can solve as we would for any other linear equation: x =  7 x = 1 What's the next step? Multiply by 7 On Your Own: Solve for x: 6 (x) = 18 6(x) = 18 Simplify before solving. Divide both sides by 6 to isolate the x. x = 8
33 Practice (Use the positive roots).0 & Solve for x: = 9 x = 7. Solve for x: 64 = 8x x = 1. Solve for x: x = 7 x = 4. Solve for x: 11 = x x = Solve for x: = _x x = Solve for x: 4x² = 100 x = 5 7. Solve for x: 169 = 6x x = 1 8. x = 15 x = 5 9. x² = 16 1 x = x = 81 x = 18 9
34 Unit Quiz: The following questions appeared on the CAHSEE If x = 7, then x = A. 7 B. 1 7 C. 7 1 D. 7 Standard.0. Which of the following is equivalent to 4(x + 5)  (x + ) = 14 A. 4x x  6 = 14 B. 4x x + 6 = 14 C. 4x x + = 14 D. 4x x  = 14 Standard 4.0. Which of the following is equivalent to the equation shown below: 0 x 4 = x 5 A. x(x 5) = 80 B. 0 (x  5) = 4x C. 0x = 4(x  5) D. 4 = x + (x  5) Standard Questions 1 and 5 are based on concepts taught in Unit 1. 0
35 4. Colleen solved the equation (x + 5) = 8 using the following steps: Given: (x + 5) = 8 Step 1: 4x + 10 = 8 Step : 4x =  Step : x = ½ To get from Step to Step, Colleen A. divided both sides by 4 B. subtracted 4 from both sides C. added 4 to both sides D. multiplied both sides by 4 Standard The perimeter, P, of a square may be found by using the formula ( 1 ) P = A 4, where A is the area of the square. What is the perimeter of the square with an area of 6 square inches? A. 9 inches B. 1 inches C. 4 inches D. 7 inches Standard.0 1
36 Unit : Solving Linear Inequalities 4.0 & 5.0 An inequality is a relationship between two numbers or expressions in which one number is greater than (>) or less than (<) the other. For the most part, solving linear inequalities is very similar to solving linear equations. Let's solve two similar problems, first as an equation and secondly as an inequality: Equation Example: Solve the equality: x 7 =  x = x = 4 Apply rule of opposites: Add 7 to both sides. Inequality Example: Solve the inequality: x 7 <  x < x < 4 Apply rule of opposites: Add 7 to both sides. Let's look at another example of an inequality: Example: Solve the inequality 5x < 10 5x 10 < Apply rule of reciprocals: Divide both sides by x <  Exception: There is one important way in which inequalities differ from equations: If you multiply by a negative number, you must reverse the inequality sign. x Example: Solve the inequality > 7 x () < () (7) We apply rule of reciprocals: Multiply both sides by  (and reverse the sign). x < 14
37 On Your Own: Solve the following inequalities. 4.0 & x + (  x) > 1 6x +  x > 1 5x > 10 x >. x + 5 < 19 x + 5 < 19  x < x > = 4. x + 5 4x  1 x + 5 4x  1 x 17 X 17
38 4. 5c  47(c + ) c 4.0 & 5.0 5c  47c  1 c c  5 c 5 c 5. 4(x  ) < (x + 6) 4x  1 < 6x x < 6 x < 5 ' 6. (x  4) > x + 5 x + 8 > x + 5 x >  x < 1 4
39 Unit Quiz: The following questions appeared on the CAHSSEE. 1. Which of the following is equivalent to 1  x > (x  )? A. 1 x > x B. 1 x > x 5 C. 1 x > x 6 D. 1 x > x 7 Standard 4.0. Which of the following is equivalent to 9  x > 4(x  1)? A. 1 < 11x B. 1 > 11x C. 10 > 11x D. 6 x > 0 Standard 4.0. Solve for x: 5(x ) 6x < 9 A. x < 1.5 B. x < 1.5 C. x < D. x < 6 Standard 5.0 5
40 Mixed Review: Equations and Inequalities 4.0 & Solve for x: x 9 x = 15 Tutors: You may need to explain to students that they can rewrite x 9 15 this problem as = and then just cross multiply, as they x 1 have learned. x = 1. Solve for x: 6(x  7) = 6(4x  ) x = . 4x  = (x  5) x = 6
41 4. x = x 4 x = Solve for x: (4x  )  x < 18 x < 6. x  > 15 x > 6 Tutors: For more practice, as well as a more indepth lesson on solving simple linear equations, refer to the Algebra & Functions workbook. 7
42 .0 Unit 4: Equations & Inequalities Involving Absolute Value On the CAHSEE, you will be asked to solve algebraic equations and inequalities that involve absolute value. Before we look at these kinds of problems, let's do a quick review of absolute value. A. Review of Absolute Value Definition: The absolute value of a number is its distance from 0. This distance is always expressed as a positive number, regardless of whether the number is positive or negative. It is easier to understand this by examining a number line: 4 = 4 4 = 4 Notice that the distance of both numbers from 0 is 4 units. Note: Distance is never negative! The absolute value of 4, expressed as 4, is 4 because it is 4 units from zero. The absolute value of 4, expressed as 4, is also 4 because it is 4 units from zero. On the other hand, any number between absolute value bars has two possible values: a positive value and a negative value: Example: = or  Note: While the absolute value is +, the number between the absolute value bars can be + or : or  8
43 On Your Own: Find the absolute value..0 Absolute Value ,00 1,00 1,00 1, ½ ½ If x = 8, what is x? 8 or 8 (Hint: There are answers.) 9
44 B. Finding the Absolute Value of an Expression.0 Rule: To find the absolute value of an expression, simplify the expression between the absolute value bars and take the absolute value of the result. Example: = 7 = 7 Example: = 7 = 7 On Your Own: Solve within the absolute value bars; then find the absolute value of the expression (i.e. simplify the expression). The first one has been done for you =  =  16 = 14 = = 48 = = 0 = = 147 = = = = = = = 5 Now we are ready to solve algebraic equations involving absolute value. 40
45 C. Algebraic Equations Involving Absolute Value.0 Rule: To solve an equation involving absolute value, make two separate equations (one positive and one negative) and solve each. Example: x + = 7 Clear the absolute value bars and split the equation into two cases (positive and negative): (x + ) = 7 (x + ) = 7 Don t forget: Everything within the bars will be multiplied by 1. Remove the parentheses in each equation: x + = 7 x  = 7 Solve each equation: x + = 7 x = 5 x  = 7 x = 9 x = 9 Answer: The solution set for x + = 7 is {9, 5}. Note: Absolute value problems will usually have two values for x. Now check by substituting each answer in the original equation: 5 + = 7 = = 7 = 7 41
46 On Your Own.0 Example: x  14 = 9 Clear the absolute value bars and split the equation into two cases (positive and negative): (x  14 ) = 9  (x 14) = 9 Remove the parentheses in each equation: x  14 = 9 x + 14 = 9 Solve each equation: x  14 = 9 x = x + 14 = 9 x = 5 x = 5 Answer: The solution set for x  14 = 9 is {5, }. Check by substituting each answer in the original equation:  14 = = 99 = 9 4
47 Practice: Find the solution set for each equation x  5 = x = x = x = 5 x = 4 x = 46. x x = 8 x = 9 x = x = 9 x = x = x + 8 = 14 x = x = 11 4
48 6. x = 1.0 x = 18 x = x = 1 x = 18 x = x + x = 11 x = 6 x = x + x = 11 x = 6 x = 16 1 = x + 4 = 11 x = 4 x = 18 Now we are ready to look at inequalities involving absolute value. 44
49 D. Inequalities Involving Absolute Value.0 There are two different rules for solving inequalities involving absolute value: one for "less than" inequalities and one for "greater than" inequalities. i. Less than and "Less than or Equal to" Inequalities Rule: For problems in form < #, follow this pattern: # < <+# Example: Assume x is an integer and solve for x: x+ < 6 Clear the absolute value bars according to the pattern:  6 < x + < 6 Isolate the x: 6  < x +  < 6  We apply the rule of opposites 9 < x < We combine like terms 9 < x < We divide both sides by to isolate x. Answer: We write the solution to x+ < 6 as { 9, }. This means that x is greater than 9 and less than. We can see this on a number line; the solution set for x includes everything between the two black circles: Check your solution. Pick a number between into the original equation: (4)+ < 6 9 and. Plug it Does the inequality hold true? Yes: 5 < 6 5 < 6 45
50 Integer Only Solutions.0 On the CAHSEE, you may be asked to find a solution that involves integers only. Integers are positive or negative whole numbers. Fractions and decimals are not integers. (Note: 0 is an integer.) Example: If k is an integer, what is the solution to x  < 1? Clear the absolute value bars according to the pattern: 1 < x  < 1 Isolate the x: 1 < x  < < x  + < 1 + Rule of Opposites 1 < x < We combine like terms The solution set consists of one integer: {} Now check your solution in the original problem: x  < 1?  < 1? 0 < 1 Yes, it checks out! 46
51 On Your Own:.0 Example: Assume x is an integer & solve for x: x  4 < 11 Clear the absolute value bars according to the pattern:  11 < x 4 < 11 Isolate and solve for x: 7 < x < 5 Check your solution in the original equation by setting x equal to any number within the solution set: x  4 < 11 ( ) 4 < 11 x  4 < 11 ( ) 4 < 11 () 4 < < < < 11 True statement (4) 4 < < 11 8 < 11 8 < 11 True statement 47
52 Practice: Find the solution to each inequality x + 5 < 914 < x < 4. 5x  14 < 11 < x < 5 5. x + < 54 < x < 1 Solution: {, , 1, 0} x < 1 6 > x >  8 or 6 > x > 5. x x or x Now let's look at "greater than" inequalities: 48
53 ii. Greater than Inequalities.0 Rule: For problems in form > #, split the inequality into two separate cases: < # and > +# Flip the inequality sign and change the sign of the #. Example: Assume x is an integer and solve for x: x > 5. Split the inequality into two cases: x < 5 x >5 Clear the absolute value bars: x < 5 x > 5 Isolate and solve for the x in each case: x + < 5 + x <  x < x < 1 x + > 5 + x > 8 x 8 > x > 4 Answer: x is less than 1 or x is greater than 4: Check your work. Plug in a number less than 1: Plug in a number greater than 4: () > 5 (5) > 5 Does it check out? Yes: 7 > 5 Does it check out? Yes: 7 > 5 49
54 On Your Own.0 Example: Assume x is an integer and solve for x: x + > 5 Split into inequalities: x + < 5 x + >5 Clear the absolute value bars: x + < 5 x + > 5 Isolate the x in each case: x +  < 5  x +  > 5  = x < 8 = x > x 8 < x > x < 4 x > 1 Check your work by testing different values for x in the original inequality: x + > 5. Does the inequality hold true? Are your values consistent with the solution? For x < 4, students can set x equal to 5, 6, 7, and so on. They might also set it equal to 4 and  to show that the inequality does not hold for values greater than or equal to 4. For x > 1, student can set x equal to,, 4, and so on. They might also set it equal to 1 and 0 to show that the inequality does not hold for values less than or equal to 1. 50
55 On Your Own.0 Example: Assume x is an integer and solve for x: x + 4 > 6. Split the inequality into two cases: x + 4 < 6 x + 4 > 6 Clear the absolute value bars: x + 4 < 6 x + 4 > 6 Isolate the x in each case: x + 4 < 6 x <  10 x < 5 x + 4 > 6 x > x > 1 Shade the number line where the inequality is true: Tutor version 51
56 Practice: Solve each inequality. Remember to follow all steps x  > 6. x  <  6 x <  x  > 6 x > x > x <  6 x <  x < x > 6 x > 0 x > 10. x + 11 > 7. x + 11 <  7 x < 18 x < 6 x + 11 > 7 x > 4 x > x + > 6 x + <  6 x < 9 x + > 6 x > 5
57 5. x 5 > X 5 <  x < x  5 > x > x 1 > x  1 <  x <  x >1 Flip sign ( ) x  1 > x > 4 x<  Flip sign ( ) For problem 6, shade the number line where the inequality is true: Now check your solution in the original equation by setting x equal to different numbers within the solution set: x 1 > x 1 > () 1 > () 1 > 4 1 > 6 1 > 5 > 5 > 5 > True statement 5 > True statement 5
58 Unit Quiz: The following questions appeared on the CAHSEE. 1. If x is an integer, which of the following is the solution set for the following inequality: x = 15? A. {0, 5} B. {5, 5} C. {5, 0, 5} D. {0, 45} Standard.0. Assume k is an integer and solve for k k > 4 A. {, , 1, 1,, } B. {, , 1, 0, 1, } C. {, 1, 0, 1, } D. {, 1, 1,, } Standard.0. If x is an integer, what is the solution to x < 1? A. {} B. {, , 1, 0, 1} C. {} D. {1, 0, 1,, } Standard.0 4. Assume y is an integer and solve for y. y+ = 9 A. {11, 7} B. {7, 7} C. {7, 11} D. {11, 11} Standard.0 54
59 Unit 5: Solving Systems of Equations Algebraically 9.0 On the CAHSEE, you will be given a "system of equations." This just means that instead of getting one equation with one unknown variable, such as x + 5 = 7, you will be given two equations and two unknown variables. Example of a system of equations: x + y = 15 x  y = 1 Note: To solve for 1 variable, we only need 1 equation. However, to solve for variables, we need equations! There are two methods for solving systems of equations. Let's begin with the first method: Method 1: Substitution In this method, we begin with one equation and one variable, and solve in terms of the second variable. Example: Steps: We need to choose one equation to solve first, and to pick one variable to solve for. We could start with either one, but the first equation is simpler: x  y = 0 We will solve for x in terms of y: x  y = 0. x  y + y = 0 + y x = y Add y to both sides. 55
60 We can now plug in the value of x, which is y, in the second equation and then solve for y: 9.0 x + y = 15 (y) + y= 15 Plug in the value of x: y Note: Now we have only one variable & can easily solve for it. 4y + y = 15 5y = 15 5y 15 = 5 5 y = We get rid of the parentheses. We add like terms. We divide both sides by 5. This is our answer. Now that we have solved for y, we can solve for x by plugging the value of y into the first equation: x  y = 0 x  () = 0 x  6 = 0 x = 6 Plug in the value of y: We can check our answers by plugging in the values of both x and y into either equation to see if the equation holds true: x + y = 15 (6) + ()= 15 Plug in values for x & y 1 + = 15 Do the two sides balance? Yes 56
61 Let's look at another example: 9.0 x + 5y = 11 x + 4y = 1 Let s solve the second equation first since its x coefficient (the numerical factor that appears before the variable) is 1. (Note: This is easier to solve than the first equation, whose variable has a coefficient of.) x + 4y = 1 x + 4y  4y = 14y x = 14y Subtract 4y from both sides. Now that we have a value for x, let s plug it into the first equation and solve for y: x + 5y = 11 (14y) + 5y = 11 Plug in the value of x: 14y 68y + 5y = 11 We use distributive law 6  y = 11 We combine like terms y = 15 We subtract 6 from both sides y = 5 Isolate by dividing both sides by  Now that we have a value for y, let s plug it into the second equation and solve for x. x + 4y = 1 x + 4(5) = 1 x + 0 = 1 Plug in the value of y Get rid of the parentheses x = 7 Subtract 0 from both sides Continued 57
62 9.0 Check your work: Plug the x and y values into the first equation: x + 5y = 11 (7) + 5(5) = = 11 Does this check out? Yes Example: 7x + y = 14 x  y = 15 Which equation would you start with? Second equation Which variable would you solve for first? x Explain: We can get x by itself by adding y to both sides. Example: x + y = 1 x  y = 0 Which equation would you start with? First equation Which variable would you solve for first? y Explain: We can get y by itself by subtracting x from both sides. 58
63 9.0 Let's look at one last example: Which equation would you start with here? Not clear yet. In the last two examples, we began by solving for a variable with a coefficient of 1. In this example, neither equation has a variable with a coefficient of 1. However, we can convert one of the equations so that we do have a coefficient of 1 for the x variable. Note: It's easier to work with the first equation because each term can be divided evenly by : x + 4y = 8 x 4y 8 + = x + y = 4 We can divide each term by. Now we can proceed as we did with the other examples: Solve for x: x + y = 4 x + y  y = 4  y x = 4  y We subtract y from both sides. Now that we have a value for x, let s plug it into the second equation and solve for y: x + 5y = 14 (4  y) + 5y = 14 Plug in value for x. 16y + 5y = 14 Use distributive law and solve for y. y = y =  59
64 Now that we have a value for y, plug it into the first equation and solve for x. 9.0 x + 4y = 8 x + 4() = 8 Plug in the value of y. x  8 = 8 x = 16 x = 8 Solve for x. Check your work: Plug the x and y values into the second equation: (8) + 5 () = = 14 Does this check out? Yes 60
65 On Your Own 9.0 Find the solution for the system of equations below: Choose one equation to begin with. x + y = 1 Solve for one variable in terms of the other variable. x = 1  y Now substitute the value of the variable in the other equation and solve for your second variable: (1  y) + y = y + y = 195y = 5 y = 1 Plug the value of your second variable into the equation you began with, and solve for the first variable: x + (1) = 1 x = 9 Check your work: x + y = 19 (9) + (1) = = 19 61
66 9.0 Practice: Use the method of substitution to find the solution for the systems of equations below: 1. Solve for x and y: x y = 14 x y = 0 x = 14 and y = 8. What is the solution for the system of equations? x = 7 and y = 1. What is the solution for the system of equations? x = and y = 5 6
67 Method : Adding Equations 9.0 Sometimes it is easier to eliminate one of the variables by adding the equations. The goal is to create a new equation with only one variable. Example: What is the solution for the system of equations? Notice that the x in the first equation and the +x in the second equation give us the sum of 0 because the coefficients in the two terms are opposites. In other words, we can get rid of the x variable by simply adding the two equations together. 4y  x = 10 y + x = 11 7y + 0 = 1 Make sure each term is lined up! Notice that the xvariables cancel out. Now we have one equation with one variable: 7y = 1 7y = 1 y = Divide both sides by 7 Now plug the value for y into either of the two original equations and find the value of x: 4y  x = 10 4()  x = 10 We plug in the value of y. 1  x = 10 We get rid of the parentheses x = 101 We subtract 1 from both sides. x =  Combine like terms. x = Now divide both sides by . x = 1 Does this check out? () + (1) = 11 6
68 Let's look at another example: 9.0 Example: What is the solution for the system of equations? Notice that the xcoefficients are the same in each equation:. Therefore, we cannot eliminate a variable by adding the two equations. However, if we multiply the second equation by 1, we can then add the two equations and get a 0 value for the xvariable. 1 (x + y = 4) Note: We must multiply each term by 1. This is equal to x y = 4. Now we can add the two equations: x + y = 10 x  y =  4 Line up each term in the equations and add. 0  y = 6 Notice that the xvariables canceled out. Now we have one equation in one variable: y = 6 Let's now solve the equation: y 6 = y =  Divide both sides by . Now plug the value for y into either equation and find the value of x: x + () = 10 x  = 10 We plug in the value of y. We get rid of the parentheses. x  + = 10 + We add to both sides. x = 1 What should you do now? Divide both sides by. x 1 = x = 1 1 = 6 Check: x + y = 10 ( 1 ) + () = = 10 64
69 Rules for Solving Systems of Equations with Addition 9.0 When we add equations, our goal is to eliminate one of the variables so that we can solve one equation with one variable. In order to eliminate a variable, the coefficients (the numbers in front of that variable) must be opposites (e.g.  and +). When we add the two equations, we will get a 0 value for that variable. As we saw on the previous page, whenever the coefficients are the same, we can multiply each term in one equation by 1. We will then have opposite coefficients and when we add the two equations, we will get a 0 value for that variable. Exercise: Look at each of the following systems of equations. Write which variable can be eliminated (using addition). If we need to first multiply an equation by 1, indicate this. The y variable can be eliminated First, multiply one equation by 1; then we can add the equations and eliminate the y variable. First multiply one equation by 1; then we can add the equations and eliminate the x variable. Multiply 1 equation by 1 and eliminate x variable Eliminate y variable by adding the equations. 65
70 9.0 Let's look at another example: We can eliminate the x if we first multiply one equation by 1. Then, the coefficients for the xvariable in the two equations (1 and 1) will be opposites: x  y = 15 x  y = 151(x + y = 10)  x  y = 10 Now we can add the two equations: x  y = 15 + (x  y = 10) 0x  5y = 55y = 5 Now we can solve for y: 5y = 5 5y 5 = 5 5 y = 1 We divide both sides by 5. Plug the value for y into either equation and find the value of x: x + y = 10 x + (1) = 10 x  = 10 x = 1 Plug in the value for y. Get rid of the parentheses. Add to both sides. Check by plugging in the x and y values into the other equation: x  y = 15 (1)  (1) = = 15 66
71 On Your Own: Use the method of addition or subtraction to solve the system of equations below: x + y = 6 4x y = 15 Add or subtract the equations and eliminate a variable: 9.0 x + y = 6 + 4x  y = 15 7x + 0 = 1 Solve for the other variable: 7x = 1 x = Plug in the value of the variable (from step ) in either equation and solve for the remaining variable: 4()  y = y = 15 y = Check your work by plugging the values of both variables in the remaining equation: () + ( ) = = 6 67
72 Practice: Use the addition method to solve: Find the solution to the system of equations. x = 1 and y = 1. Find the solution to the system of equations. x = and y = 68
73 Let's examine one last example: 9.0 Neither term can be eliminated through addition. However, we can create an equivalent equation in which one of the terms can be eliminated. Look at the second equation: x  y = 9 If we multiply the entire equation by, we get the following: (x  y = 9) = 9x  y = 7 We now have two y variables with opposite coefficients: + and . We can now add the two equations and eliminate the y variable: 4x + y = 1 + 9x  y = 7 Add the equations. 1x + 0y = 9 1x = 9 We can now solve for x: 1x = 9 1x 9 = 1 1 x = Divide both sides by 1. We can now plug the x value into either of the two original equations and solve for y. The second equation is easiest: ()  y = 9 Plug in the value of x. 9  y = 9 Get rid of parentheses. y = 0 Solve for y. Plug the x and y values into the first equation to see if they hold true: 4() + (0) = 1 Does it check out? Yes 69
74 9.0 Practice: Use the method of addition to find the solution to each system of equations: 1. Find the solution to the system of equations. First, we will multiply the first equation by : (x + y = 5) 4x + y = 10 Now add the two equations: 4x + y = 10 + x  y = 11 7x = 1 x = 4() + y = 10 y = 1. Find the solution to the system of equations: First, we will multiply the first equation by : (5x + y = 15) 10x  6y = 0 Now add the two equations: 10x  6y = x + 4y = 0 y = 50 y = 5 10x + 4(5) = 0 10x = 10 x = 1 70
75 9.0 Mixed Practice: Use either method (substitution or addition) to find the solution to each system of equations. 1. Find the solution to the system of equations. x = 1 and y = . Find the solution to the system of equations. x = 6 and y = 4. Find the solution to the system of equations. x = 5 and y =  4. Find the solution to the system of equations: x = 4 and y = 6 71
76 Unit Quiz: The following problems appeared on the CAHSEE. 1. What is the solution of the system of equations shown below? y = x 5 y = x A. (1, ) B. (1, ) C. (5, 10) D. (5, 10) Standard 9.0. What is the solution of the system of equations shown below? 7x + y = 84x y = 6 A. (, ) B. (, ) C. (, ) D. (,) Standard 9.0 7
77 . Which graph represents the system of equations shown below? y = x + y = x + Answer: B Standard: 9.0 7
78 Unit 6: Monomials, Binomials, & Polynomials 10.0 On the CAHSEE, you will be given problems in which you must add, subtract, multiply and divide monomial, binomial and polynomial expressions. Let's begin with a review of the vocabulary. A. Vocabulary Review Term: A term, which is the basic unit of an algebraic expression, is a specific value. A term can be a number: Examples: A term can be the product of a number and one or more variables: Examples of Terms: 4x Product of 4 and x x 1 Product of x and 5ab Product of 5, a, and b Coefficient: A coefficient is the numerical factor of the term; it is the number that appears before the variable. Example: In the term xy, the coefficient is : The above term also includes two variables: x and y. Note: All terms have a numerical coefficient, even if it is not written. Example: x² This term consists of the variable x². Notice that there is no number in front of the variable. It is understood that we have 1x²; in other words, the numerical coefficient is 1. We do not need to write the coefficient. What is the coefficient of x²? 1 74
79 10.0 Monomial Expression: An expression consisting of one term Examples: 4x z² 18abc xy ab x Addition or subtraction signs connect separate terms. Example: x + 4 1st Term + nd Term Binomial Expression: An expression consisting of two terms connected by the plus (+) or minus () sign. Examples: x² + 4 x x + x 5 4x + 4x² + x abc + a 5c Trinomial Expression: An expression consisting of three terms connected by the plus (+) or minus () sign. Examples: x² + x + 5 4x² + x x²  5x
80 10.0 Polynomial Expression: The prefix poly means "many." A polynomial expression is a numerical expression that consists of many terms; as with all expressions, each term is connected by a plus (+) or minus () sign. The term polynomial is usually used for four or more terms. Example: 4x + x + 5x + Degrees of Polynomials Look at the polynomial expression: x + x + x Notice that each term contains a different degree (exponent) of x. There are special names for each of these terms: 1. Terms that have an exponent of 1 are called linear terms. Example: 5x The exponent (1) doesn't need to be written.. Terms that have an exponent of are called quadratic terms. Example: x² Only x is raised to the nd power (not ). Terms that have an exponent of are called cubic terms. Example: 5x Only x is raised to the rd power (not 5) Note: The entire expression is named by the highest degree and the number of terms in the expression: Example: x + x² + x is a cubic trinomial 76
81 B. Adding & Subtracting Polynomials 10.0 Polynomials are expressions that contain multiple terms. Example: x 4 + x + 5x + 6x  8 The above polynomial expression has five terms. i. Adding Polynomials To add polynomials, add like terms. Let's solve a problem together. Example: x² + 4x + + x² + x + Rewrite the problem so that all like terms are next to one another: x² + x² + 4x + x + + Note: Like terms have the same variable and are raised to the same power. Now add like terms: x² + x² + 4x + x + + = 5x² + 7x + 5 Note: To add like terms, we add the coefficients of the like terms. The variables and exponents do not change! On Your Own: 1. 6x + x² + 4x x + 14x² + x x + 16x + 7x +. 1a + 9a² + 4a a + 6a² + 1a + 9 a + 15a + 6a
82 ii. Subtracting Polynomials 10.0 To subtract polynomials, combine like terms. Example: x² + 4x + (x² + x + ) Get rid of the parentheses. Be sure to subtract the entire expression: pay attention to the sign of each term that is subtracted. x² + 4x + (x² + x + ) = x² + 4x + x² x Rewrite the problem so that all like terms are next to one another: x² + 4x +  x²  x = x²  x² + 4x  x +  Now combine like terms: x²  x² + 4x  x +  = x² + x + 1 On Your Own: 1. 14x² + 9x + 15 (5x²  6x + 19) 9x + 15x y²  17y + 1 (1y² + 16y  1) y  y
83 C. Multiplying Monomials by Monomials 10.0 In the example below, we must multiply the two monomials in order to simplify the expression. Example: (5a b )(ab ) Multiply the coefficients (the numbers before the variables). 5  = 15 Multiply similar variables (i.e. variables that have a common base). When multiplying variables that have a common base: add the exponents and keep the base.) Imagine an exponent of 1 here! a² a = a² + ¹ = a b² b = b² + = b Multiply all terms: 15 a b = 15a b On Your Own: (x²)(8x³) =  8 x + = 4x 5 (x²)(7y²) = ()(7) x²y² = 14x²y² 5a b 48a b 5 = (5)(8) a + b = 40a 5 b 9 (x²y)(7xy²) = ()(7) x + 1 y 1+ = 1x y 79
84 D. Dividing Monomials by Monomials 10.0 In the example below, we must divide the two monomials in order to simplify the expression. 1x y Example: 5 4xy The fraction bar means division! Divide the coefficients: Divide out common factors is a factor of both 1 and 4 so = 4 8 Divide similar variables (i.e. variables that have a common base). When dividing variables that have a common base, keep the base and subtract the exponent in the denominator from the exponent in the numerator. x = x  ¹ = x² x y = y¹  1 = y = 4 y 5 y Multiply all terms. Remember, when dealing with fractions, we first multiply numerator by numerator; we then multiply denominator by denominator. Your answer should be a simpler fraction than that in the original problem: 7 1 x² 4 8 y 7 x = 4 8 y 1 7x = 4 8y Note: No variable appearing in the numerator should appear in the denominator and viseversa. If the same variable appears in both the numerator and the denominator, you aren't finished solving the problem: It can be simplified further. 80
85 On Your Own: x y x y 14 = x 418 y 55 = 7 x 9. 1xy 18x y = y x or y x 5 4 9x y. 4 18x y = xy 7 8 8a y a y = 4 7 a 5 y 16ab 5. 5 = 4a b a b 81
86 E. Multiplying Monomials by Binomials 10.0 On the CAHSEE, you may be asked to multiply a monomial expression by a binomial expression: Example: x(x + 4) This expression contains both... a monomial (x) and a binomial (x + 4) To multiply monomials by binomials, use the distributive law. Distributive Law of Multiplication: For any three numbers a, b, and c, a (b + c) = (a x b) + (a x c) = ab + ac In other words, multiply the monomial by each term in the binomial. Example: x(x + 4) = (x x) + (x 4) = x² + 1x On Your Own: 1. 4x(x  8) = 1x²  x x 1 x 1 =x = x². x²(x  5) = 6x  15x. 4x(4x² + 6x) = 16x  4x 4. 7x² (4x  11) = 8x  77x 8
87 F. Multiplying Monomials by Trinomials 10.0 Just as we used the distributive law to multiply a monomial by a binomial (in the last lesson), we use the distributive law to multiply a monomial by a trinomial. Example: x(x² + x  5) This expression contains both... a monomial (x) and a trinomial (x² + x  5) Apply the distributive law of multiplication: Example: x(x² + x  5) = (x x²) + (x x) + x(5) = x + 6x²  15x Note: Each term in the answer has different variables: x x² x Therefore, the terms cannot be combined, and the expression cannot be further simplified. Example: 4x²(x²y + 4xy² + 5xy) = (4x²)(x²y) + (4x²)(4xy²) + (4x²)(5xy) = 1x 4 y  16x³y²  0x y Notice that each term in the answer has different variables: x 4 y x³y² x y Therefore, the terms cannot be combined (added), and the expression cannot be further simplified. 8
88 On Your Own: x²(x²  14x  18) = 6x 4 + 8x + 6x. x(1x  8x + 0) = 6x 44x + 60x. 5x²(10x²  1x + 7) = 50x 460x + 5x 4. 6x(8x 5 + 7x 46x) = 48x 6 + 4x 56x 5. 7x²(x²  7x  8) = 14x x + 56x 6. 5xy(x²y²  6xy  8x + 9y) = 5x y  0x y  40x y + 45xy 7. 8x²y(6xy² + 7xy  9x + 8) = 48x y  56x y + 7x y  64x y 8. 6x²(6x²  1y + 7x + 9) = 6x 4 + 7x y  4x  54x 84
89 G. Multiplying Binomials by Binomials 10.0 When multiplying two binomial expressions, use the distributive law to multiply each term in the first expression by each term in the second expression. One way to keep track of all the terms that need to be multiplied is to use the FOIL method. FOIL stands for First, Outer, Inner, Last. That is the order in which we multiply the terms. Example: ( + 7x)(6 + x) Apply the FOIL method: 6 = 18 Multiply the first term in each expression: Multiply the outer term in each expression: x = 6x Multiply the inner term in each expression: 7x 6 = 4x Multiply the last term in each expression: 7x x = 14x² Now add all four answers and combine the like terms: 14x² + 6x + 4x + 18= 14x² + 48x + 18 Note: We end with a trinomial expression. We always write trinomials from highest degree terms to lowest degree terms. 85
90 On Your Own 10.0 Example: (x + )(x + ) = Use the FOIL method to solve the above problem. First: x² Outer: x Inner: x Last: 6 Add and combine like terms: x² + x + x + 6 = x² + 5x + 6 Note: We can also multiply two binomials by using a grid. Look at the example below. Example: (x + )(x + ) Now add all the terms and combine any like terms: x² + 5x
91 On Your Own 10.0 Example: Simplify (x  )(x  ) = First: x x = x² Outer: x () = x Inner:  x = x Last:   = 4 Now add and combine like terms: x²  x  x + 4 = x²  4x
92 On Your Own 10.0 Example: (x )(x +4) Use FOIL: First: x x = x Outer: x 4 = 4x Inner:  x = 6x Last:  4 = 1 Combine: x  x  1 Example: (x + 8)(x  8) Use FOIL: First: x x = x Outer: x 8 = 8x Inner: 8 x = 8x Last: 88 = 64 Combine: x
93 Practice: (x + 5)(x ) 6x + 11x (x + 4)² Hint: Write without exponent: (x + 4)(x + 4) x² + 8x (y )² 9y²  18y (x )² x²  6x (x + y)(x  y) x² + y² 6. The length of the rectangle is 8 units longer than its width. Find the area of the rectangle. x + 8 x x² + 8x 89
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