CAHSEE on Target UC Davis, School and University Partnerships


 Juniper Richardson
 2 years ago
 Views:
Transcription
1
2
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:
4
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
CAHSEE on Target UC Davis, School and University Partnerships
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,
More informationMath 0980 Chapter Objectives. Chapter 1: Introduction to Algebra: The Integers.
Math 0980 Chapter Objectives Chapter 1: Introduction to Algebra: The Integers. 1. Identify the place value of a digit. 2. Write a number in words or digits. 3. Write positive and negative numbers used
More information3. Power of a Product: Separate letters, distribute to the exponents and the bases
Chapter 5 : Polynomials and Polynomial Functions 5.1 Properties of Exponents Rules: 1. Product of Powers: Add the exponents, base stays the same 2. Power of Power: Multiply exponents, bases stay the same
More informationSTUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS
STUDY GUIDE FOR SOME BASIC INTERMEDIATE ALGEBRA SKILLS The intermediate algebra skills illustrated here will be used extensively and regularly throughout the semester Thus, mastering these skills is an
More information1.3 Algebraic Expressions
1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts,
More informationAnswers to Basic Algebra Review
Answers to Basic Algebra Review 1. 1.1 Follow the sign rules when adding and subtracting: If the numbers have the same sign, add them together and keep the sign. If the numbers have different signs, subtract
More informationUnit 3 Polynomials Study Guide
Unit Polynomials Study Guide 75 Polynomials Part 1: Classifying Polynomials by Terms Some polynomials have specific names based upon the number of terms they have: # of Terms Name 1 Monomial Binomial
More informationMATH 65 NOTEBOOK CERTIFICATIONS
MATH 65 NOTEBOOK CERTIFICATIONS Review Material from Math 60 2.5 4.3 4.4a Chapter #8: Systems of Linear Equations 8.1 8.2 8.3 Chapter #5: Exponents and Polynomials 5.1 5.2a 5.2b 5.3 5.4 5.5 5.6a 5.7a 1
More informationMATH 60 NOTEBOOK CERTIFICATIONS
MATH 60 NOTEBOOK CERTIFICATIONS Chapter #1: Integers and Real Numbers 1.1a 1.1b 1.2 1.3 1.4 1.8 Chapter #2: Algebraic Expressions, Linear Equations, and Applications 2.1a 2.1b 2.1c 2.2 2.3a 2.3b 2.4 2.5
More informationAlum Rock Elementary Union School District Algebra I Study Guide for Benchmark III
Alum Rock Elementary Union School District Algebra I Study Guide for Benchmark III Name Date Adding and Subtracting Polynomials Algebra Standard 10.0 A polynomial is a sum of one ore more monomials. Polynomial
More informationMATH Fundamental Mathematics II.
MATH 10032 Fundamental Mathematics II http://www.math.kent.edu/ebooks/10032/funmath2.pdf Department of Mathematical Sciences Kent State University December 29, 2008 2 Contents 1 Fundamental Mathematics
More informationPlacement Test Review Materials for
Placement Test Review Materials for 1 To The Student This workbook will provide a review of some of the skills tested on the COMPASS placement test. Skills covered in this workbook will be used on the
More informationAlgebra Revision Sheet Questions 2 and 3 of Paper 1
Algebra Revision Sheet Questions and of Paper Simple Equations Step Get rid of brackets or fractions Step Take the x s to one side of the equals sign and the numbers to the other (remember to change the
More informationWhat are the place values to the left of the decimal point and their associated powers of ten?
The verbal answers to all of the following questions should be memorized before completion of algebra. Answers that are not memorized will hinder your ability to succeed in geometry and algebra. (Everything
More informationEquations and Inequalities
Rational Equations Overview of Objectives, students should be able to: 1. Solve rational equations with variables in the denominators.. Recognize identities, conditional equations, and inconsistent equations.
More informationexpression is written horizontally. The Last terms ((2)( 4)) because they are the last terms of the two polynomials. This is called the FOIL method.
A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are
More informationCD 1 Real Numbers, Variables, and Algebraic Expressions
CD 1 Real Numbers, Variables, and Algebraic Expressions The Algebra I Interactive Series is designed to incorporate all modalities of learning into one easy to use learning tool; thereby reinforcing learning
More informationEXPONENTS. To the applicant: KEY WORDS AND CONVERTING WORDS TO EQUATIONS
To the applicant: The following information will help you review math that is included in the Paraprofessional written examination for the Conejo Valley Unified School District. The Education Code requires
More informationAlgebra Cheat Sheets
Sheets Algebra Cheat Sheets provide you with a tool for teaching your students notetaking, problemsolving, and organizational skills in the context of algebra lessons. These sheets teach the concepts
More informationName Date Block. Algebra 1 Laws of Exponents/Polynomials Test STUDY GUIDE
Name Date Block Know how to Algebra 1 Laws of Eponents/Polynomials Test STUDY GUIDE Evaluate epressions with eponents using the laws of eponents: o a m a n = a m+n : Add eponents when multiplying powers
More information( 7) + 4 = (9) =  3 ( 3) + 7 = ( 3) = 2
WORKING WITH INTEGERS: 1. Adding Rules: Positive + Positive = Positive: 5 + 4 = 9 Negative + Negative = Negative: ( 7) + ( 2) =  9 The sum of a negative and a positive number: First subtract: The answer
More informationOperations with Algebraic Expressions: Multiplication of Polynomials
Operations with Algebraic Expressions: Multiplication of Polynomials The product of a monomial x monomial To multiply a monomial times a monomial, multiply the coefficients and add the on powers with the
More informationx 41 = (x²)²  (1)² = (x² + 1) (x²  1) = (x² + 1) (x  1) (x + 1)
Factoring Polynomials EXAMPLES STEP 1 : Greatest Common Factor GCF Factor out the greatest common factor. 6x³ + 12x²y = 6x² (x + 2y) 5x  5 = 5 (x  1) 7x² + 2y² = 1 (7x² + 2y²) 2x (x  3)  (x  3) =
More informationx n = 1 x n In other words, taking a negative expoenent is the same is taking the reciprocal of the positive expoenent.
Rules of Exponents: If n > 0, m > 0 are positive integers and x, y are any real numbers, then: x m x n = x m+n x m x n = xm n, if m n (x m ) n = x mn (xy) n = x n y n ( x y ) n = xn y n 1 Can we make sense
More informationLESSON 6.2 POLYNOMIAL OPERATIONS I
LESSON 6.2 POLYNOMIAL OPERATIONS I Overview In business, people use algebra everyday to find unknown quantities. For example, a manufacturer may use algebra to determine a product s selling price in order
More informationPreAlgebra Interactive Chalkboard Copyright by The McGrawHill Companies, Inc. Send all inquiries to:
PreAlgebra Interactive Chalkboard Copyright by The McGrawHill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGrawHill 8787 Orion Place Columbus, Ohio 43240 Click the mouse button
More informationPrep for College Algebra
Prep for College Algebra This course covers the topics shown below. Students navigate learning paths based on their level of readiness. Institutional users may customize the scope and sequence to meet
More informationThis assignment will help you to prepare for Algebra 1 by reviewing some of the things you learned in Middle School. If you cannot remember how to complete a specific problem, there is an example at the
More informationMAT 0950 Course Objectives
MAT 0950 Course Objectives 5/15/20134/27/2009 A student should be able to R1. Do long division. R2. Divide by multiples of 10. R3. Use multiplication to check quotients. 1. Identify whole numbers. 2. Identify
More informationMth 95 Module 2 Spring 2014
Mth 95 Module Spring 014 Section 5.3 Polynomials and Polynomial Functions Vocabulary of Polynomials A term is a number, a variable, or a product of numbers and variables raised to powers. Terms in an expression
More informationSuccessful completion of Math 7 or Algebra Readiness along with teacher recommendation.
MODESTO CITY SCHOOLS COURSE OUTLINE COURSE TITLE:... Basic Algebra COURSE NUMBER:... RECOMMENDED GRADE LEVEL:... 811 ABILITY LEVEL:... Basic DURATION:... 1 year CREDIT:... 5.0 per semester MEETS GRADUATION
More informationMonomials with the same variables to the same powers are called like terms, If monomials are like terms only their coefficients can differ.
Chapter 7.1 Introduction to Polynomials A monomial is an expression that is a number, a variable or the product of a number and one or more variables with nonnegative exponents. Monomials that are real
More informationA Quick Algebra Review
1. Simplifying Epressions. Solving Equations 3. Problem Solving 4. Inequalities 5. Absolute Values 6. Linear Equations 7. Systems of Equations 8. Laws of Eponents 9. Quadratics 10. Rationals 11. Radicals
More informationCopy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any.
Algebra 2  Chapter Prerequisites Vocabulary Copy in your notebook: Add an example of each term with the symbols used in algebra 2 if there are any. P1 p. 1 1. counting(natural) numbers  {1,2,3,4,...}
More informationPolynomial Expression
DETAILED SOLUTIONS AND CONCEPTS  POLYNOMIAL EXPRESSIONS Prepared by Ingrid Stewart, Ph.D., College of Southern Nevada Please Send Questions and Comments to ingrid.stewart@csn.edu. Thank you! PLEASE NOTE
More informationDate: Section P.2: Exponents and Radicals. Properties of Exponents: Example #1: Simplify. a.) 3 4. b.) 2. c.) 3 4. d.) Example #2: Simplify. b.) a.
Properties of Exponents: Section P.2: Exponents and Radicals Date: Example #1: Simplify. a.) 3 4 b.) 2 c.) 34 d.) Example #2: Simplify. a.) b.) c.) d.) 1 Square Root: Principal n th Root: Example #3: Simplify.
More informationSECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS
(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1 SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES Be able to identify polynomial, rational, and algebraic
More informationUNIT 5 VOCABULARY: POLYNOMIALS
2º ESO Bilingüe Page 1 UNIT 5 VOCABULARY: POLYNOMIALS 1.1. Algebraic Language Algebra is a part of mathematics in which symbols, usually letters of the alphabet, represent numbers. Letters are used to
More informationModuMath Algebra Lessons
ModuMath Algebra Lessons Program Title 1 Getting Acquainted With Algebra 2 Order of Operations 3 Adding & Subtracting Algebraic Expressions 4 Multiplying Polynomials 5 Laws of Algebra 6 Solving Equations
More informationAnswer Key. Algebra 1. Practice Test. Algebra 1 Practice Test. Copyright Karin Hutchinson, All rights reserved.
Algebra 1 Practice Test Answer Key Copyright Karin Hutchinson, 2011. All rights reserved. Please respect the time, effort, and careful planning spent to prepare these materials. The distribution of this
More information2.3. Finding polynomial functions. An Introduction:
2.3. Finding polynomial functions. An Introduction: As is usually the case when learning a new concept in mathematics, the new concept is the reverse of the previous one. Remember how you first learned
More informationVocabulary Words and Definitions for Algebra
Name: Period: Vocabulary Words and s for Algebra Absolute Value Additive Inverse Algebraic Expression Ascending Order Associative Property Axis of Symmetry Base Binomial Coefficient Combine Like Terms
More informationIntroduction to the Practice Exams
Introduction to the Practice Eams The math placement eam determines what math course you will start with at North Hennepin Community College. The placement eam starts with a 1 question elementary algebra
More information( yields. Combining the terms in the numerator you arrive at the answer:
Algebra Skillbuilder Solutions: 1. Starting with, you ll need to find a common denominator to add/subtract the fractions. If you choose the common denominator 15, you can multiply each fraction by one
More informationObjectives. By the time the student is finished with this section of the workbook, he/she should be able
QUADRATIC FUNCTIONS Completing the Square..95 The Quadratic Formula....99 The Discriminant... 0 Equations in Quadratic Form.. 04 The Standard Form of a Parabola...06 Working with the Standard Form of a
More informationFlorida Math 0028. Correlation of the ALEKS course Florida Math 0028 to the Florida Mathematics Competencies  Upper
Florida Math 0028 Correlation of the ALEKS course Florida Math 0028 to the Florida Mathematics Competencies  Upper Exponents & Polynomials MDECU1: Applies the order of operations to evaluate algebraic
More informationChapter 3. Algebra. 3.1 Rational expressions BAa1: Reduce to lowest terms
Contents 3 Algebra 3 3.1 Rational expressions................................ 3 3.1.1 BAa1: Reduce to lowest terms...................... 3 3.1. BAa: Add, subtract, multiply, and divide............... 5
More informationI N D U BRIDGING THE GAP TO A/S LEVEL MATHS C T I O N
I N D U BRIDGING THE GAP TO A/S LEVEL MATHS C T I O N INTRODUCTION TO A LEVEL MATHS The Mathematics Department is committed to ensuring that you make good progress throughout your A level or AS course.
More informationGreatest Common Factor (GCF) Factoring
Section 4 4: Greatest Common Factor (GCF) Factoring The last chapter introduced the distributive process. The distributive process takes a product of a monomial and a polynomial and changes the multiplication
More informationUnderstand the difference between linear equations and quadratic equations Multiply polynomials Factor quadratic equations
LESSON 26: Quadratic Equations part 1 Weekly Focus: quadratic equations Weekly Skill: factoring Lesson Summary: For the warmup, students will solve a problem about oil usage. Activity 1 is an introduction
More informationYear 9 set 1 Mathematics notes, to accompany the 9H book.
Part 1: Year 9 set 1 Mathematics notes, to accompany the 9H book. equations 1. (p.1), 1.6 (p. 44), 4.6 (p.196) sequences 3. (p.115) Pupils use the Elmwood Press Essential Maths book by David Raymer (9H
More informationALGEBRA I A PLUS COURSE OUTLINE
ALGEBRA I A PLUS COURSE OUTLINE OVERVIEW: 1. Operations with Real Numbers 2. Equation Solving 3. Word Problems 4. Inequalities 5. Graphs of Functions 6. Linear Functions 7. Scatterplots and Lines of Best
More informationName Intro to Algebra 2. Unit 1: Polynomials and Factoring
Name Intro to Algebra 2 Unit 1: Polynomials and Factoring Date Page Topic Homework 9/3 2 Polynomial Vocabulary No Homework 9/4 x In Class assignment None 9/5 3 Adding and Subtracting Polynomials Pg. 332
More informationFactoring and Applications
Factoring and Applications What is a factor? The Greatest Common Factor (GCF) To factor a number means to write it as a product (multiplication). Therefore, in the problem 48 3, 4 and 8 are called the
More informationFlorida Math for College Readiness
Core provides a fourthyear math curriculum focused on developing the mastery of skills identified as critical to postsecondary readiness in math. This fullyear course is aligned with Florida's Postsecondary
More informationMath Help and Additional Practice Websites
Name: Math Help and Additional Practice Websites http://www.coolmath.com www.aplusmath.com/ http://www.mathplayground.com/games.html http://www.ixl.com/math/grade7 http://www.softschools.com/grades/6th_and_7th.jsp
More informationExponents and Radicals
Exponents and Radicals (a + b) 10 Exponents are a very important part of algebra. An exponent is just a convenient way of writing repeated multiplications of the same number. Radicals involve the use of
More informationSummer Mathematics Packet Say Hello to Algebra 2. For Students Entering Algebra 2
Summer Math Packet Student Name: Say Hello to Algebra 2 For Students Entering Algebra 2 This summer math booklet was developed to provide students in middle school an opportunity to review grade level
More informationAlgebra 1A and 1B Summer Packet
Algebra 1A and 1B Summer Packet Name: Calculators are not allowed on the summer math packet. This packet is due the first week of school and will be counted as a grade. You will also be tested over the
More informationMyMathLab ecourse for Developmental Mathematics
MyMathLab ecourse for Developmental Mathematics, North Shore Community College, University of New Orleans, Orange Coast College, Normandale Community College Table of Contents Module 1: Whole Numbers and
More informationUnit 3: Algebra. Date Topic Page (s) Algebra Terminology 2. Variables and Algebra Tiles 3 5. Like Terms 6 8. Adding/Subtracting Polynomials 9 12
Unit 3: Algebra Date Topic Page (s) Algebra Terminology Variables and Algebra Tiles 3 5 Like Terms 6 8 Adding/Subtracting Polynomials 9 1 Expanding Polynomials 13 15 Introduction to Equations 16 17 One
More information2.1 Algebraic Expressions and Combining like Terms
2.1 Algebraic Expressions and Combining like Terms Evaluate the following algebraic expressions for the given values of the variables. 3 3 3 Simplify the following algebraic expressions by combining like
More informationNSM100 Introduction to Algebra Chapter 5 Notes Factoring
Section 5.1 Greatest Common Factor (GCF) and Factoring by Grouping Greatest Common Factor for a polynomial is the largest monomial that divides (is a factor of) each term of the polynomial. GCF is the
More informationPractice Math Placement Exam
Practice Math Placement Exam The following are problems like those on the Mansfield University Math Placement Exam. You must pass this test or take MA 0090 before taking any mathematics courses. 1. What
More informationChapter 5  Polynomials and Polynomial Functions
Math 233  Spring 2009 Chapter 5  Polynomials and Polynomial Functions 5.1 Addition and Subtraction of Polynomials Definition 1. A polynomial is a finite sum of terms in which all variables have whole
More informationFlorida Math for College Readiness
Core Florida Math for College Readiness Florida Math for College Readiness provides a fourthyear math curriculum focused on developing the mastery of skills identified as critical to postsecondary readiness
More informationActually, if you have a graphing calculator this technique can be used to find solutions to any equation, not just quadratics. All you need to do is
QUADRATIC EQUATIONS Definition ax 2 + bx + c = 0 a, b, c are constants (generally integers) Roots Synonyms: Solutions or Zeros Can have 0, 1, or 2 real roots Consider the graph of quadratic equations.
More informationtroduction to Algebra
Chapter Three Solving Equations and Problem Solving Section 3.1 Simplifying Algebraic Expressions In algebra letters called variables represent numbers. The addends of an algebraic expression are called
More informationSometimes it is easier to leave a number written as an exponent. For example, it is much easier to write
4.0 Exponent Property Review First let s start with a review of what exponents are. Recall that 3 means taking four 3 s and multiplying them together. So we know that 3 3 3 3 381. You might also recall
More informationBasic Math Refresher A tutorial and assessment of basic math skills for students in PUBP704.
Basic Math Refresher A tutorial and assessment of basic math skills for students in PUBP704. The purpose of this Basic Math Refresher is to review basic math concepts so that students enrolled in PUBP704:
More informationCOGNITIVE TUTOR ALGEBRA
COGNITIVE TUTOR ALGEBRA Numbers and Operations Standard: Understands and applies concepts of numbers and operations Power 1: Understands numbers, ways of representing numbers, relationships among numbers,
More information(2 4 + 9)+( 7 4) + 4 + 2
5.2 Polynomial Operations At times we ll need to perform operations with polynomials. At this level we ll just be adding, subtracting, or multiplying polynomials. Dividing polynomials will happen in future
More information3.1. RATIONAL EXPRESSIONS
3.1. RATIONAL EXPRESSIONS RATIONAL NUMBERS In previous courses you have learned how to operate (do addition, subtraction, multiplication, and division) on rational numbers (fractions). Rational numbers
More informationMATH REFRESHER I. Notation, Solving Basic Equations, and Fractions. Rockefeller College MPA Welcome Week 2016
MATH REFRESHER I Notation, Solving Basic Equations, and Fractions Rockefeller College MPA Welcome Week 2016 Lucy C. Sorensen Assistant Professor of Public Administration and Policy 1 2 Agenda Why Are You
More informationAlgebra 1 Unit 1 Days 1 and 2.notebook. July 13, Algebra 1. Unit 1. Relationships between Quantities and Expressions. Day 1.
Algebra 1 Unit 1 Relationships between Quantities and Expressions Day 1 Jul 12 7:55 PM 1 What Are Our Learning Goals? MGSE9 12.A.SSE.1 Interpret expressions that represent a quantity in terms of its context.
More informationEQUATIONS. Main Overarching Questions: 1. What is a variable and what does it represent?
EQUATIONS Introduction to Variables, Algebraic Expressions, and Equations (2 days) Overview of Objectives, students should be able to: Main Overarching Questions: 1. Evaluate algebraic expressions given
More informationTable of Contents Sequence List
Table of Contents Sequence List 368102215 Level 1 Level 5 1 A1 Numbers 010 63 H1 Algebraic Expressions 2 A2 Comparing Numbers 010 64 H2 Operations and Properties 3 A3 Addition 010 65 H3 Evaluating
More informationFlorida Math Correlation of the ALEKS course Florida Math 0022 to the Florida Mathematics Competencies  Lower and Upper
Florida Math 0022 Correlation of the ALEKS course Florida Math 0022 to the Florida Mathematics Competencies  Lower and Upper Whole Numbers MDECL1: Perform operations on whole numbers (with applications,
More informationAddition and Multiplication of Polynomials
LESSON 0 addition and multiplication of polynomials LESSON 0 Addition and Multiplication of Polynomials Base 0 and Base  Recall the factors of each of the pieces in base 0. The unit block (green) is x.
More information1.3 Polynomials and Factoring
1.3 Polynomials and Factoring Polynomials Constant: a number, such as 5 or 27 Variable: a letter or symbol that represents a value. Term: a constant, variable, or the product or a constant and variable.
More informationPreAlgebra Curriculum Map 8 th Grade Unit 1 Integers, Equations, and Inequalities
Key Skills and Concepts Common Core Math Standards Unit 1 Integers, Equations, and Inequalities Chapter 1 Variables, Expressions, and Integers 12 days Add, subtract, multiply, and divide integers. Make
More informationCourse Title: Math A Elementary Algebra
Course Title: Math A Elementary Algebra Unit 0: Course Introduction In this unit you will get familiar with important course documents, such as syllabus and communication policy. You will register on MyMathLab
More informationMATH LEVEL 1 ARITHMETIC (ACCUPLACER)
MATH LEVEL ARITHMETIC (ACCUPLACER) 7 Questions This test measures your ability to perform basic arithmetic operations and to solve problems that involve fundamental arithmetic concepts. There are 7 questions
More informationAccuplacer Elementary Algebra Study Guide for Screen Readers
Accuplacer Elementary Algebra Study Guide for Screen Readers The following sample questions are similar to the format and content of questions on the Accuplacer Elementary Algebra test. Reviewing these
More informationDefinition 8.1 Two inequalities are equivalent if they have the same solution set. Add or Subtract the same value on both sides of the inequality.
8 Inequalities Concepts: Equivalent Inequalities Linear and Nonlinear Inequalities Absolute Value Inequalities (Sections 4.6 and 1.1) 8.1 Equivalent Inequalities Definition 8.1 Two inequalities are equivalent
More informationAlgebra 1 Chapter 3 Vocabulary. equivalent  Equations with the same solutions as the original equation are called.
Chapter 3 Vocabulary equivalent  Equations with the same solutions as the original equation are called. formula  An algebraic equation that relates two or more reallife quantities. unit rate  A rate
More informationMath 10C. Course: Polynomial Products and Factors. Unit of Study: Step 1: Identify the Outcomes to Address. Guiding Questions:
Course: Unit of Study: Math 10C Polynomial Products and Factors Step 1: Identify the Outcomes to Address Guiding Questions: What do I want my students to learn? What can they currently understand and do?
More informationUNIT TWO POLYNOMIALS MATH 421A 22 HOURS. Revised May 2, 00
UNIT TWO POLYNOMIALS MATH 421A 22 HOURS Revised May 2, 00 38 UNIT 2: POLYNOMIALS Previous Knowledge: With the implementation of APEF Mathematics at the intermediate level, students should be able to: 
More informationRadicals  Multiply and Divide Radicals
8. Radicals  Multiply and Divide Radicals Objective: Multiply and divide radicals using the product and quotient rules of radicals. Multiplying radicals is very simple if the index on all the radicals
More informationSUNY 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)
More informationName: Date: Algebra 2/ Trig Apps: Simplifying Square Root Radicals. Arithmetic perfect squares: 1, 4, 9,,,,,,...
RADICALS PACKET Algebra 2/ Trig Apps: Simplifying Square Root Radicals Perfect Squares Perfect squares are the result of any integer times itself. Arithmetic perfect squares: 1, 4, 9,,,,,,... Algebraic
More informationChapter 4  Decimals
Chapter 4  Decimals $34.99 decimal notation ex. The cost of an object. ex. The balance of your bank account ex The amount owed ex. The tax on a purchase. Just like Whole Numbers Place Value  1.23456789
More informationMATH 90 CHAPTER 6 Name:.
MATH 90 CHAPTER 6 Name:. 6.1 GCF and Factoring by Groups Need To Know Definitions How to factor by GCF How to factor by groups The Greatest Common Factor Factoring means to write a number as product. a
More informationPark Forest Math Team. Meet #5. Algebra. Selfstudy Packet
Park Forest Math Team Meet #5 Selfstudy Packet Problem Categories for this Meet: 1. Mystery: Problem solving 2. Geometry: Angle measures in plane figures including supplements and complements 3. Number
More informationUnit 1: Polynomials. Expressions:  mathematical sentences with no equal sign. Example: 3x + 2
Pure Math 0 Notes Unit : Polynomials Unit : Polynomials : Reviewing Polynomials Epressions:  mathematical sentences with no equal sign. Eample: Equations:  mathematical sentences that are equated with
More information1.1 Solving a Linear Equation ax + b = 0
1.1 Solving a Linear Equation ax + b = 0 To solve an equation ax + b = 0 : (i) move b to the other side (subtract b from both sides) (ii) divide both sides by a Example: Solve x = 0 (i) x = 0 x = (ii)
More informationChapter R  Basic Algebra Operations (69 topics, due on 05/01/12)
Course Name: College Algebra 001 Course Code: R3RK6CTKHJ ALEKS Course: College Algebra with Trigonometry Instructor: Prof. Bozyk Course Dates: Begin: 01/17/2012 End: 05/04/2012 Course Content: 288 topics
More informationSelfDirected Course: Transitional Math Module 2: Fractions
Lesson #1: Comparing Fractions Comparing fractions means finding out which fraction is larger or smaller than the other. To compare fractions, use the following inequality and equal signs:  greater than
More informationAlgebra I. Copyright 2014 Fuel Education LLC. All rights reserved.
Algebra I COURSE DESCRIPTION: The purpose of this course is to allow the student to gain mastery in working with and evaluating mathematical expressions, equations, graphs, and other topics, with an emphasis
More information