Algebra II Practice. Dr. Barbara Sandall, Ed.D. Travis Olson, M.S.

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BY Dr. Barbara Sandall, Ed.D. Dr. Melfried Olson, Ed.D. Travis Olson, M.S. COPYRIGHT 2006 Mark Twain Media, Inc. ISBN 978-1-58037-753-9 Printing No. 404043-EB Mark Twain Media, Inc., Publishers Distributed by Carson-Dellosa Publishing Company, Inc. The purchase of this book entitles the buyer to reproduce the student pages for classroom use only. Other permissions may be obtained by writing Mark Twain Media, Inc., Publishers. This product has been correlated to state, national, and Canadian provincial standards. Visit www.carsondellosa.com to search and view its correlations to your standards. All rights reserved. Printed in the United States of America.

Table of Contents Table of Contents Introduction to the Series...1 Common Mathematics Symbols and Terms...2 Algebra Rules and Laws...12 Chapter 1: Solving Equations and Problems...13 Simplifying Expressions and Solving Equations With One Variable...13 Changing Words Into Symbols; Problem Solving With Equations...16 Chapter 2: Inequalities...23 Inequalities...23 Graphing Inequalities...24 Solving Inequalities...24 Working With Absolute Values...26 Chapter 3: Linear Equations and Inequalities...32 Linear Equations and Graphs...32 Linear Inequalities...38 Linear Systems...40 Chapter 4: Polynomial Products and Factors...46 Simplifying Polynomials...46 Laws of Exponents...46 Multiplying and Factoring Polynomials..49 Solving Polynomial Equations...52 Chapter 5: Rational Expressions...55 Rational Expressions...55 Scientific Notation...58 Chapter 6: Roots, Radicals, and Complex Numbers...64 Simplifying Radicals, Products, Quotients, Sums, Differences...64 Simplifying Binomials With Radicals and Solving Radical Equations...70 Decimal Representation and Complex Numbers...75 Chapter 7: Quadratic Equations and Functions...81 Solving Quadratic Equations...81 Quadratic Functions and Graphs...89 Chapter 8: Variation...95 Direct Variation, Proportion, Inverse Variation, Joint Variation...95 Algebra II Check-up...100 Practice, Challenge, and Checking Progress Answer Keys...109 Check-up Problems Answer Keys...123 References...126 iii

Basic Overview: Simplifying Expressions and Solving Equations With One Variable Algebraic expressions can be simplified by applying the order of operations, particularly when the expressions contain multiple operations. (1) First evaluate within parentheses from innermost to outermost using rules 2, 3, and 4 in order; (2) Evaluate all exponents; (3) Multiply and/or divide from left to right; and then (4) Add and/or subtract, also from left to right. Equations can sometimes be solved by the guess-and-check method, but more often their solutions follow directly from using the principles of algebra. Equations can be systematically solved, as follows: (1) Use the distributive property to remove parentheses and simplify each side of the equation; (2) Apply the addition property of equality to variables on one side of the equation and constants on the other side; and (3) Apply the multiplication property of equality to isolate the variable. Examples of Simplifying Expressions and Solving Equations With One Variable Example of Simplifying Algebraic Expressions: 3 (4 + 5) = 3 (9) = 27 Order of Operations Example of Simplifying Equations With One Variable by Addition and Subtraction: x 4 = 9 (x 4) + 4 = 9 + 4 Addition Property of Equality x + (-4 + 4) = 13 Associative Property of Addition x = 13 Identity Property of Addition Example of Simplifying Equations With One Variable by Multiplication and Division: 7b = 5(6 + 1) 7b = 35 7 7 b = 5 13

(cont.) Practice: Simplifying Expressions and Solving Equations with One Variable Directions: Simplify the following expressions. 1. 7x 2 + 3(x 9) + 5 2. 40 4 3(9 2) 3. 2 2 5(x + 3) 4. 2(x 5) + 3(x + 11) x 5. 3x 6(2 x) + 3(x 2) + 8 6. a 2a [3a (4a 5)] 7. 3 3 a 2 4 a + 7a 2(9 a) 8. 13(5 t) (5 t) 12(5 t) Directions: Simplify and solve the following equations. Work the problem on your own paper, if you need more room. 9. 2a + 3a 13 = a + 3 10. 3(x 55) = 0 11. 3(x 55) + 22 = x + 55 12. 5 2x + 9 3 = 3x + 2 13. If 8x + 5(3 + x) a = 15 + 5x, then a =? 14. 3x + 4(x 3) = 6x 9 15. 5t (4 t) = 8 (2 + t) 14

(cont.) 16. 17(2x + 3) = 15(2x + 3) 17. 3x + 4x = 6(2x 1) 18. 18 4 2 (1 3x) = 98 19. 3 5 (x + 9) = 3 20. 12 + [5 (9 + 2x)] = 7 5x Challenge Problems: Simplifying Expressions and Solving Equations with One Variable Directions: Work the problems on your own paper, and write the answers on the lines below. 1. Simplify. 2x {2 [2(x 2) (2 x)]} 2. If 3x + 5(x 2) + a = 9x + 13, then a =? 3. Simplify. 3[5 2 5(14 7) + 9 0 8] 4. Simplify and solve. 22x + 9(2 3x) = 3(2 3x) 5. Simplify and solve. 43(19 7x) + 32(19 7x) = 25(7x 19) 15

(cont.) Basic Overview: Changing Words Into Symbols; Problem Solving With Equations Word problems describe relationships between or among number ideas and specific arithmetic operations. We often transform verbal problems into algebraic expressions quite naturally. To support this process, practice the following steps: (1) Read the problem carefully to identify the quantities you are being asked to find; (2) If possible, draw a sketch to help visualize the problem; (3) Choose a variable and use it to express the unknown quantities; (4) Write the equation that represents the problem statement; (5) Solve the equation and answer the questions being asked; and (6) Check your solution in your original equation. Examples of Changing Words Into Symbols; Problem Solving With Equations Example of Changing Word Phrases Into Algebraic Expressions: A number is increased by 5 A number is replaced with the letter a a + 5 Example of Changing Sentences Into Algebraic Expressions: If a runner runs at x miles/hour, what would his speed be if he was 2 miles per hour slower? Speed = x 2 miles an hour slower is represented by -2 x + (-2) = x 2 Example of Changing Scientific Formulas Into Algebraic Equations: The amount of work done is represented by the formula: work = force times distance or W = f d. How much work was done if a force of 25 kg was moved a distance of 6 m? W = work done Force f = 25 kg Distance d = 6 m W = 25 kg 6 m 16