Algebra in 15 Minutes a Day (Junior Skill Builders)
Post on 27-Dec-2016
Embed Size (px)
Junior Skill Builders
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page i
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page ii
N E W Y O R K
Junior Skill Builders
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page iii
Copyright 2009 LearningExpress, LLC.
All rights reserved under International and Pan-American CopyrightConventions. Published in the United States by LearningExpress, LLC, New York.
Library of Congress Cataloging-in-Publication Data:Junior skill builders: algebra in 15 minutes a day.
p. cm. ISBN 978-1-57685-673-4 1. AlgebraStudy and teaching (Middle school) 2. AlgebraStudy and
teaching (Secondary) I. LearningExpress (Organization) II. Title: Algebrain fifteen minutes a day.
Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
For more information or to place an order, contact LearningExpress at:2 Rector Street26th FloorNew York, NY 10006
Or visit us at:www.learnatest.com
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page iv
S E C T I O N 1 : A L G E B R A BA S I C S 2 1
Lesson 1: Explaining Variables and Terms 23 Learning the language of algebra Explaining like and unlike terms
Lesson 2: Adding and Subtracting 29 Combining double signs Adding and subtracting like and unlike terms
Lesson 3: Multiplying and Dividing 37 Multiplying and dividing like and unlike terms
Lesson 4: Single-Variable Expressions 47 Order of operations Evaluating expressions that contain one variable
Lesson 5: Multivariable Expressions 57 Simplifying and evaluating multivariable expressions
Lesson 6: Algebra Word Expressions 67 Translating words into algebraic expressions
C O N T E N T S
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page v
v i con t en t s
Lesson 7: Factoring 73 Factoring single-variable and multivariable expressions
Lesson 8: Exponents 81 Working with positive, negative, and fractional exponents
S E C T I O N 2 : S O LV I N G A N D G R A P H I N G E Q UAT I O N S A N D I N E Q UA L I T I E S 8 9
Lesson 9: Solving Single-Step Algebraic Equations 91 Using the four basic operations to solve single-step equations
Lesson 10: Solving Multistep Algebraic Equations 99 Using the four basic operations to solve multistep equations
Lesson 11: Radicals 109 Working with radicals and fractional exponents Solving equations with radicals
Lesson 12: Slope-Intercept Form 117 Defining slope-intercept form Putting equations in slope-intercept form Finding parallel and perpendicular lines
Lesson 13: Input/Output Tables 125 Finding the rule and the missing value in input/output tables
Lesson 14: Graphing Equations 135 Learning the coordinate plane Graphing the equation of a line
Lesson 15: Finding Equations from Graphs 145 Finding the equation of a line from the graph of a line Determining if an ordered pair is on a line
Lesson 16: Distance 153 Finding the distance between two points by counting, using
the Pythagorean theorem, and using the distance formula
Lesson 17: Functions, Domain, and Range 161 Learning about functions, domain, and range Determining if an equation is a function: the vertical line test Finding the domain and range of a function
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page vi
contents v i i
Lesson 18: Systems of EquationsSolving with Substitution 171 Using substitution to solve systems of equations with two or
Lesson 19: Systems of EquationsSolving with Elimination 181 Using elimination to solve systems of equations with two or
Lesson 20: Algebraic Inequalities 189 Solving single-variable and multivariable inequalities
Lesson 21: Graphing Inequalities 197 Graphing single inequalities Graphing the solution set of a system of inequalities
Lesson 22: Polynomials and FOIL 207 Describing algebraic expressions and equations by name Multiplying two binomials using FOIL
Lesson 23: Quadratic Factoring 213 Factoring a trinomial into two binomials: reverse FOIL Factoring a binomial into two binomials by finding the
difference of perfect squares
Lesson 24: Quadratic Equation 223 Solving a quadratic equation using factoring Solving a quadratic equation using the quadratic formula
S E C T I O N 3 : U S I N G A L G E B R A 2 3 1
Lesson 25: Algebra Word Problems 233 Recognizing keywords in algebra word problems Solving real-world problems with algebra
Lesson 26: Using Algebra: Ratios and Proportions 241 Learning ratios and proportions Finding exact values using ratios and proportions
Lesson 27: Using Algebra: Statistics and Probability 247 Solving statistics problems Solving probability problems
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page vii
v i i i c on t en t s
Lesson 28: Using Algebra: Percents and Simple Interest 255 Finding percent increase and percent decrease Calculating simple interest
Lesson 29: Using Algebra: Sequences 263 Finding rules and terms of arithmetic and geometric
Lesson 30: Using Algebra: Geometry 273 Finding perimeter, area, and volume of two- and three-
Hints for Taking Standardized Tests 297
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page viii
Junior Skill Builders
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page ix
JSBA_00_i-x_FM.qxd:JSB 12/18/08 12:19 PM Page x
IF YOUVE NEVER seen algebra before, then the math you have seen hasmostly involved numbers. You may know that algebra seems to involve the let-ter x a whole lot, and that may be enough to make you want to close this book.But really, algebra is just about finding missing numbers. We do not alwaysknow the value of a quantity, so we use a letter like x to hold the place of thatvalue.
Algebra can help us find how much interest our money is earning in abank, or algebra might help us find the probability of drawing the ace of spadesfrom a deck of cards. You might think that you do not use algebra often in every-day life, but any time you ask yourself a question like, How many magazinescan I buy with the money I have? you are using algebra. The number of mag-azines you can buy is a number you do not know, and that number multipliedby the cost of one magazine must be less than or equal to the amount of moneyyou have.
I N T R O D U C T I O N
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 1
2 in t r oduc t i on
USING THIS BOOK
This book is divided into 30 lessons, plus a pretest and a posttest. The lessonsare designed to be completed in 15 minutes. Each day, you can learn a little alge-bra without taking too much of your time.
THE BOOK AT A GLANCE
Each lesson in this book builds on the previous lesson. We start with an expla-nation of algebra and the words that are often used when we are learning alge-bra. What is a term? What is a variable? What is a constant? If you already knowthe answers to these questions, you are off to a fast start. If not, do not worry;that is what this book is forto introduce you to algebra and show you how touse it to solve a variety of problems.
Before the lessons begin, well start with a pretest. This 30-question test willgive you an idea of how well you know algebra now. Every question containsan answer explanation and lets you know from which lesson in the book thequestion came. Remember, it is only a pretestif you knew the answer to everyquestion, you would not be reading this book!
After you have been working through all 30 lessons, the book ends with aposttest. The posttest, like the pretest, is 30 questions. Compare your results onthe posttest to your pretest. Did you correctly answer the types of questions youstruggled with in the pretest? Are there a few topics that you might need tospend more time reviewing? You can always go back and reread a lesson or two.Each lesson contains plenty of practice problems, with complete answer expla-nations for every question, plus tips to help you remember how to solve aproblem or another way to go about solving problems.
The pretest is nextgrab a pen or pencil and get started!
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 2
THIS PRETEST WILL test your knowledge of basic algebra. The 30 questionsare presented in the same order that the topics they cover are presented in thisbook. The book builds skills from lesson to lesson, so skills you have learned inearlier lessons may be required to solve problems found in later lessons. Thismeans that you may find the problems toward the beginning of the test easierto solve than the problems toward the end of the test.
For each of the 30 multiple-choice questions, circle the correct answer. Thepretest will show you which algebra topics are your strengths and which top-ics you might need to reviewor learn for the first time. Explanations are pro-vided for every question, along with the lesson number that teaches the skillsneeded to answer the question. After taking the pretest, you might realize thatyou need only to review a few lessons. Or you might benefit from workingthrough the book from start to finish.
Save your answers after you have completed the pretest. When you havefinished the lessons in this book, take the posttest. Compare your pretest to yourposttest to see how you have improved.
P R E T E S T
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 3
4 pre t e s t
1. 12q6 + 4q6 =a. 8q6
b. 16c. 16q6
2. (45a4b9c5) (9ab3c3) = a. 5a3b6c2
3. What is 10 + 6j when j = 3?a. 18b. 8c. 8d. 28
4. Simplify 7g6 + 9h + 2h 8g6.a. 4g6hb. 2g6 4hc. 5g6 + hd. 15g6 + 11h
5. Which of the following is nine less than three times a number?a. x 9b. 9 3xc. 3x 9d. 3(x 9)
6. Which of the following is equal to 12x7 24x3?a. 12x4
b. 12x3(x4 2)c. 12x(x7 2x3)d. 12x(x6 2x)
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 4
7. (4w9)3 =a. 4w12
8. What is the value of z in z 7 = 9?a. 2b. 1c. 2d. 16
9. If 4n = 28m, what is n in terms of m?a. 32mb. 22mc. 7md. 7m
10. If 7k 11 = 10, what is the value of k?a. 3b. 1c. 2d. 21
11. What is the value of p in p6 + 13 = p 2?
a. 6b. 12c. 15d. 18
12. Which of the following is equivalent to ?a. o3
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 5
6 pre t e s t
13. Find the slope of the equation 8y = 16x 4.a. 12b. 2c. 8d. 16
14. What equation was used to build the input/output table below?
a. y = 3x + 4b. y = 4x 1c. y = 5x 2d. y = 7x
15. Which of the following is the graph of y = 4x 3?a. y
x 5 4 3 2 1 1 2 3 4 5
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 6
5 4 3 2 1 1 2 3 4 5
x 5 4 3 2 1 1 2 3 4 5
5 4 3 2 1 1 2 3 4 5
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 7
8 pre t e s t
16. What is the equation of the line graphed here?
a. y = x + 7b. y = x 7c. y = x 7d. y = x + 7
17. Find the distance from (3,3) to (9,13).a. 12 unitsb. 16 unitsc. 20 unitsd. 24 units
18. Which of the following is NOT a function?a. y = x7
b. x = y2
c. x = d. y = |x|
19. What is the solution to the following system of equations?4y + x = 73y 2x = 30a. x = 9, y = 4b. x = 3, y = 1c. x = 3, y = 12d. x = 6, y = 6
5 4 3 2 1 1 2 3 4 5
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 8
20. What inequality is represented by the following graph?
a. x < 4b. x > 4c. x 4d. x 4
21. The solution set of what system of inequalities is shown as the darkerregion on the following graph?
a. y > 14x, y 4xb. y < 14x, y 4xc. y > 14x, y 4xd. y < 14x, y 4x
22. (7x 1)(x + 2) =a. 7x2 2b. 7x2 + 13x 2c. 7x2 + 14x 2d. 7x2 14x 2
23. Which of the following shows x2 + 10x + 24 correctly factored?a. (x + 5)(x + 5)b. (x + 3)(x + 8)c. (x + 12)(x + 2)d. (x + 4)(x + 6)
5 4 3 2 1 1 2 3 4 5
0 1 2 3 4 5 6 7 8 9 1010 9 8 7 6 5 4 3 2 1
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 9
10 pre t e s t
24. Find the roots of x2 + 6x = 7.a. x = 7, x = 1b. x = 7, x = 1c. x = 7, x = 1d. x = 7, x = 1
25. Lindsays bowling score is seven less than three times Jamies score. If Lindsays score is 134, what is Jamies score?a. 39b. 41c. 43d. 47
26. The ratio of yellow beads to green beads on a necklace is 3:10. If there are 60yellow beads on the necklace, how many green beads are on the necklace?a. 70b. 100c. 200d. 600
27. The mean of a set of seven values is 15. If six of the values are 9, 19, 17, 16,11, and 20, what is the seventh value?a. 11b. 13c. 15d. 16
28. What is 48 after a 55% increase?a. 26.4b. 30.9c. 69.6d. 74.4
29. What is the value of the ninth term of the following arithmetic sequence?3x + 1, x 7, 2x 3, 3x + 1, . . .a. 53b. 61c. 64d. 75
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 10
30. The width of a rectangle is four times its length. If the perimeter of therectangle is 560 meters, what is the width of the rectangle?a. 56 mb. 112 mc. 224 md. 448 m
1. c. Each term has a base of q and an exponent of 6, so the base and expo-nent of your answer is q6.Add the coefficients: 12 + 4 = 16, 12q6 + 4q6 = 16q6.For more on this skill, review Lesson 2.
2. a. Divide the coefficients: 45 9 = 5. Carry the bases (a, b, and c) fromthe dividend into the answer. There are no bases in the divisor that arenot in the dividend.Subtract the exponent of a in the divisor from the exponent of a in thedividend: 4 1 = 3. The exponent of a in the answer is 3.Subtract the exponent of b in the divisor from the exponent of b in thedividend: 9 3 = 6. The exponent of b in the answer is 6.Subtract the exponent of c in the divisor from the exponent of c in thedividend: 5 3 = 2. The exponent of c in the answer is 2.
(45a4b9c5) (9a2b3c3) = 5a3b6c2
For more on this skill, review Lesson 3.3. b. Replace j with 3:
10 + 6(3)Multiply before adding:
6(3) = 18The expression becomes 10 + 18.
Add: 10 + 18 = 8.For more on this skill, review Lesson 4.
4. d. This expression has two g6 terms and two h terms.Combine the g6 terms: 7g6 8g6 = 15g6.Combine the h terms: 9h + 2h = 11h.7g6 + 9h + 2h 8g6 simplifies to 15g6 + 11h.For more on this skill, review Lesson 5.
JSBA_01_001-088.qxd:JSB 12/18/08 11:44 AM Page 11
12 pre t e s t
5. c. The keyword phrase less than signals subtraction and the keyword timessignals multiplication. Three times a number is 3x. Nine less than thatquantity is 3x 9.For more on this skill, review Lesson 6.
6. b. These terms are unlike, so they cannot be combined, but they can befactored.The factors of 12x7 are 1, 2, 3, 4, 6, 12, and x7, and the factors of 24x3 are1, 2, 3, 4, 6, 8, 12, 24, and x3 (and their negatives).Both terms have 12 and the variable x as common factors. The smallerexponent of x between the two terms is 3, so we can factor 12x3 out ofboth terms.Divide both terms by 12x3: 12x7 12x3 = x4, and 24x3 12x3 = 2.12x7 24x3 factors into 12x3(x4 2).For more on this skill, review Lesson 7.
7. d. 4w9 is raised to the third power.Raise 4 to the third power and raise w9 to the third power. 43 = 64.To find (w9)3, multiply the exponents: (9)(3) = 27, (w9)3 = w27, and (4w9)3 = 64w27.For more on this...