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Elementary Algebra

2.4 Use a General Strategy to Solve Linear Equations

Elementary Algebra2.4 Use a General Strategy to Solve Linear Equations
  1. Preface
  2. 1 Foundations
    1. Introduction
    2. 1.1 Introduction to Whole Numbers
    3. 1.2 Use the Language of Algebra
    4. 1.3 Add and Subtract Integers
    5. 1.4 Multiply and Divide Integers
    6. 1.5 Visualize Fractions
    7. 1.6 Add and Subtract Fractions
    8. 1.7 Decimals
    9. 1.8 The Real Numbers
    10. 1.9 Properties of Real Numbers
    11. 1.10 Systems of Measurement
    12. Key Terms
    13. Key Concepts
    14. Exercises
      1. Review Exercises
      2. Practice Test
  3. 2 Solving Linear Equations and Inequalities
    1. Introduction
    2. 2.1 Solve Equations Using the Subtraction and Addition Properties of Equality
    3. 2.2 Solve Equations using the Division and Multiplication Properties of Equality
    4. 2.3 Solve Equations with Variables and Constants on Both Sides
    5. 2.4 Use a General Strategy to Solve Linear Equations
    6. 2.5 Solve Equations with Fractions or Decimals
    7. 2.6 Solve a Formula for a Specific Variable
    8. 2.7 Solve Linear Inequalities
    9. Key Terms
    10. Key Concepts
    11. Exercises
      1. Review Exercises
      2. Practice Test
  4. 3 Math Models
    1. Introduction
    2. 3.1 Use a Problem-Solving Strategy
    3. 3.2 Solve Percent Applications
    4. 3.3 Solve Mixture Applications
    5. 3.4 Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem
    6. 3.5 Solve Uniform Motion Applications
    7. 3.6 Solve Applications with Linear Inequalities
    8. Key Terms
    9. Key Concepts
    10. Exercises
      1. Review Exercises
      2. Practice Test
  5. 4 Graphs
    1. Introduction
    2. 4.1 Use the Rectangular Coordinate System
    3. 4.2 Graph Linear Equations in Two Variables
    4. 4.3 Graph with Intercepts
    5. 4.4 Understand Slope of a Line
    6. 4.5 Use the Slope–Intercept Form of an Equation of a Line
    7. 4.6 Find the Equation of a Line
    8. 4.7 Graphs of Linear Inequalities
    9. Key Terms
    10. Key Concepts
    11. Exercises
      1. Review Exercises
      2. Practice Test
  6. 5 Systems of Linear Equations
    1. Introduction
    2. 5.1 Solve Systems of Equations by Graphing
    3. 5.2 Solve Systems of Equations by Substitution
    4. 5.3 Solve Systems of Equations by Elimination
    5. 5.4 Solve Applications with Systems of Equations
    6. 5.5 Solve Mixture Applications with Systems of Equations
    7. 5.6 Graphing Systems of Linear Inequalities
    8. Key Terms
    9. Key Concepts
    10. Exercises
      1. Review Exercises
      2. Practice Test
  7. 6 Polynomials
    1. Introduction
    2. 6.1 Add and Subtract Polynomials
    3. 6.2 Use Multiplication Properties of Exponents
    4. 6.3 Multiply Polynomials
    5. 6.4 Special Products
    6. 6.5 Divide Monomials
    7. 6.6 Divide Polynomials
    8. 6.7 Integer Exponents and Scientific Notation
    9. Key Terms
    10. Key Concepts
    11. Exercises
      1. Review Exercises
      2. Practice Test
  8. 7 Factoring
    1. Introduction
    2. 7.1 Greatest Common Factor and Factor by Grouping
    3. 7.2 Factor Quadratic Trinomials with Leading Coefficient 1
    4. 7.3 Factor Quadratic Trinomials with Leading Coefficient Other than 1
    5. 7.4 Factor Special Products
    6. 7.5 General Strategy for Factoring Polynomials
    7. 7.6 Quadratic Equations
    8. Key Terms
    9. Key Concepts
    10. Exercises
      1. Review Exercises
      2. Practice Test
  9. 8 Rational Expressions and Equations
    1. Introduction
    2. 8.1 Simplify Rational Expressions
    3. 8.2 Multiply and Divide Rational Expressions
    4. 8.3 Add and Subtract Rational Expressions with a Common Denominator
    5. 8.4 Add and Subtract Rational Expressions with Unlike Denominators
    6. 8.5 Simplify Complex Rational Expressions
    7. 8.6 Solve Rational Equations
    8. 8.7 Solve Proportion and Similar Figure Applications
    9. 8.8 Solve Uniform Motion and Work Applications
    10. 8.9 Use Direct and Inverse Variation
    11. Key Terms
    12. Key Concepts
    13. Exercises
      1. Review Exercises
      2. Practice Test
  10. 9 Roots and Radicals
    1. Introduction
    2. 9.1 Simplify and Use Square Roots
    3. 9.2 Simplify Square Roots
    4. 9.3 Add and Subtract Square Roots
    5. 9.4 Multiply Square Roots
    6. 9.5 Divide Square Roots
    7. 9.6 Solve Equations with Square Roots
    8. 9.7 Higher Roots
    9. 9.8 Rational Exponents
    10. Key Terms
    11. Key Concepts
    12. Exercises
      1. Review Exercises
      2. Practice Test
  11. 10 Quadratic Equations
    1. Introduction
    2. 10.1 Solve Quadratic Equations Using the Square Root Property
    3. 10.2 Solve Quadratic Equations by Completing the Square
    4. 10.3 Solve Quadratic Equations Using the Quadratic Formula
    5. 10.4 Solve Applications Modeled by Quadratic Equations
    6. 10.5 Graphing Quadratic Equations
    7. Key Terms
    8. Key Concepts
    9. Exercises
      1. Review Exercises
      2. Practice Test
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
  13. Index

Learning Objectives

By the end of this section, you will be able to:

  • Solve equations using a general strategy
  • Classify equations
Be Prepared 2.4

Before you get started, take this readiness quiz.

  1. Simplify: (a4).(a4).
    If you missed this problem, review Example 1.137.
  2. Multiply: 32(12x+20)32(12x+20).
    If you missed this problem, review Example 1.133.
  3. Simplify: 52(n+1)52(n+1).
    If you missed this problem, review Example 1.138.
  4. Multiply: 3(7y+9)3(7y+9).
    If you missed this problem, review Example 1.132.
  5. Multiply: (2.5)(6.4)(2.5)(6.4).
    If you missed this problem, review Example 1.97.

Solve Equations Using the General Strategy

Until now we have dealt with solving one specific form of a linear equation. It is time now to lay out one overall strategy that can be used to solve any linear equation. Some equations we solve will not require all these steps to solve, but many will.

Beginning by simplifying each side of the equation makes the remaining steps easier.

Example 2.37

How to Solve Linear Equations Using the General Strategy

Solve: −6(x+3)=24.−6(x+3)=24.

Try It 2.73

Solve: 5(x+3)=35.5(x+3)=35.

Try It 2.74

Solve: 6(y4)=−18.6(y4)=−18.

How To

General strategy for solving linear equations.

  1. Step 1. Simplify each side of the equation as much as possible.
    Use the Distributive Property to remove any parentheses.
    Combine like terms.
  2. Step 2. Collect all the variable terms on one side of the equation.
    Use the Addition or Subtraction Property of Equality.
  3. Step 3. Collect all the constant terms on the other side of the equation.
    Use the Addition or Subtraction Property of Equality.
  4. Step 4. Make the coefficient of the variable term to equal to 1.
    Use the Multiplication or Division Property of Equality.
    State the solution to the equation.
  5. Step 5. Check the solution. Substitute the solution into the original equation to make sure the result is a true statement.

Example 2.38

Solve: (y+9)=8.(y+9)=8.

Try It 2.75

Solve: (y+8)=−2.(y+8)=−2.

Try It 2.76

Solve: (z+4)=−12.(z+4)=−12.

Example 2.39

Solve: 5(a3)+5=−105(a3)+5=−10.

Try It 2.77

Solve: 2(m4)+3=−12(m4)+3=−1.

Try It 2.78

Solve: 7(n3)8=−157(n3)8=−15.

Example 2.40

Solve: 23(6m3)=8m23(6m3)=8m.

Try It 2.79

Solve: 13(6u+3)=7u13(6u+3)=7u.

Try It 2.80

Solve: 23(9x12)=8+2x23(9x12)=8+2x.

Example 2.41

Solve: 82(3y+5)=082(3y+5)=0.

Try It 2.81

Solve: 123(4j+3)=−17123(4j+3)=−17.

Try It 2.82

Solve: −68(k2)=−10−68(k2)=−10.

Example 2.42

Solve: 4(x1)2=5(2x+3)+64(x1)2=5(2x+3)+6.

Try It 2.83

Solve: 6(p3)7=5(4p+3)126(p3)7=5(4p+3)12.

Try It 2.84

Solve: 8(q+1)5=3(2q4)18(q+1)5=3(2q4)1.

Example 2.43

Solve: 10[38(2s5)]=15(405s)10[38(2s5)]=15(405s).

Try It 2.85

Solve: 6[42(7y1)]=8(138y)6[42(7y1)]=8(138y).

Try It 2.86

Solve: 12[15(4z1)]=3(24+11z)12[15(4z1)]=3(24+11z).

Example 2.44

Solve: 0.36(100n+5)=0.6(30n+15)0.36(100n+5)=0.6(30n+15).

Try It 2.87

Solve: 0.55(100n+8)=0.6(85n+14)0.55(100n+8)=0.6(85n+14).

Try It 2.88

Solve: 0.15(40m120)=0.5(60m+12)0.15(40m120)=0.5(60m+12).

Classify Equations

Consider the equation we solved at the start of the last section, 7x+8=−137x+8=−13. The solution we found was x=−3x=−3. This means the equation 7x+8=−137x+8=−13 is true when we replace the variable, x, with the value −3−3. We showed this when we checked the solution x=−3x=−3 and evaluated 7x+8=−137x+8=−13 for x=−3x=−3.

This figure shows why we can say the equation 7x plus 8 equals negative 13 is true when the variable x is replaced with the value negative 3. The first line shows the equation with negative 3 substituted in for x: 7 times negative 3 plus 8 might equal negative 13. Below this is the equation negative 21 plus 8 might equal negative 13. Below this is the equation negative 13 equals negative 13, with a check mark next to it.

If we evaluate 7x+87x+8 for a different value of x, the left side will not be −13−13.

The equation 7x+8=−137x+8=−13 is true when we replace the variable, x, with the value −3−3, but not true when we replace x with any other value. Whether or not the equation 7x+8=−137x+8=−13 is true depends on the value of the variable. Equations like this are called conditional equations.

All the equations we have solved so far are conditional equations.

Conditional equation

An equation that is true for one or more values of the variable and false for all other values of the variable is a conditional equation.

Now let’s consider the equation 2y+6=2(y+3)2y+6=2(y+3). Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.

.
Distribute. .
Subtract 2y2y to get the yy’s to one side. .
Simplify—the yy’s are gone! .

But 6=66=6 is true.

This means that the equation 2y+6=2(y+3)2y+6=2(y+3) is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable like this is called an identity.

Identity

An equation that is true for any value of the variable is called an identity.

The solution of an identity is all real numbers.

What happens when we solve the equation 5z=5z15z=5z1?

.
Subtract 5z5z to get the constant alone on the right. .
Simplify—the zz’s are gone! .

But 0101.

Solving the equation 5z=5z15z=5z1 led to the false statement 0=−10=−1. The equation 5z=5z15z=5z1 will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.

Contradiction

An equation that is false for all values of the variable is called a contradiction.

A contradiction has no solution.

Example 2.45

Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

6(2n1)+3=2n8+5(2n+1)6(2n1)+3=2n8+5(2n+1)

Try It 2.89

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

4+9(3x7)=−42x13+23(3x2)4+9(3x7)=−42x13+23(3x2)

Try It 2.90

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

8(13x)+15(2x+7)=2(x+50)+4(x+3)+18(13x)+15(2x+7)=2(x+50)+4(x+3)+1

Example 2.46

Classify as a conditional equation, an identity, or a contradiction. Then state the solution.

10+4(p5)=010+4(p5)=0

Try It 2.91

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 11(q+3)5=1911(q+3)5=19

Try It 2.92

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: 6+14(k8)=956+14(k8)=95

Example 2.47

Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.

5m+3(9+3m)=2(7m11)5m+3(9+3m)=2(7m11)

Try It 2.93

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

12c+5(5+3c)=3(9c4)12c+5(5+3c)=3(9c4)

Try It 2.94

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution:

4(7d+18)=13(3d2)11d4(7d+18)=13(3d2)11d

Type of equation What happens when you solve it? Solution
Conditional Equation True for one or more values of the variables and false for all other values One or more values
Identity True for any value of the variable All real numbers
Contradiction False for all values of the variable No solution
Table 2.5

Section 2.4 Exercises

Practice Makes Perfect

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, solve each linear equation.

232.

15(y9)=−6015(y9)=−60

233.

21(y5)=−4221(y5)=−42

234.

−9(2n+1)=36−9(2n+1)=36

235.

−16(3n+4)=32−16(3n+4)=32

236.

8(22+11r)=08(22+11r)=0

237.

5(8+6p)=05(8+6p)=0

238.

(w12)=30(w12)=30

239.

(t19)=28(t19)=28

240.

9(6a+8)+9=819(6a+8)+9=81

241.

8(9b4)12=1008(9b4)12=100

242.

32+3(z+4)=4132+3(z+4)=41

243.

21+2(m4)=2521+2(m4)=25

244.

51+5(4q)=5651+5(4q)=56

245.

−6+6(5k)=15−6+6(5k)=15

246.

2(9s6)62=162(9s6)62=16

247.

8(6t5)35=−278(6t5)35=−27

248.

3(102x)+54=03(102x)+54=0

249.

−2(117x)+54=4−2(117x)+54=4

250.

23(9c3)=2223(9c3)=22

251.

35(10x5)=2735(10x5)=27

252.

15(15c+10)=c+715(15c+10)=c+7

253.

14(20d+12)=d+714(20d+12)=d+7

254.

18(9r+7)=−1618(9r+7)=−16

255.

15(3r+8)=2815(3r+8)=28

256.

5(n1)=195(n1)=19

257.

−3(m1)=13−3(m1)=13

258.

114(y8)=43114(y8)=43

259.

182(y3)=32182(y3)=32

260.

248(3v+6)=0248(3v+6)=0

261.

355(2w+8)=−10355(2w+8)=−10

262.

4(a12)=3(a+5)4(a12)=3(a+5)

263.

−2(a6)=4(a3)−2(a6)=4(a3)

264.

2(5u)=−3(2u+6)2(5u)=−3(2u+6)

265.

5(8r)=−2(2r16)5(8r)=−2(2r16)

266.

3(4n1)2=8n+33(4n1)2=8n+3

267.

9(2m3)8=4m+79(2m3)8=4m+7

268.

12+2(53y)=−9(y1)212+2(53y)=−9(y1)2

269.

−15+4(25y)=−7(y4)+4−15+4(25y)=−7(y4)+4

270.

8(x4)7x=148(x4)7x=14

271.

5(x4)4x=145(x4)4x=14

272.

5+6(3s5)=−3+2(8s1)5+6(3s5)=−3+2(8s1)

273.

−12+8(x5)=−4+3(5x2)−12+8(x5)=−4+3(5x2)

274.

4(u1)8=6(3u2)74(u1)8=6(3u2)7

275.

7(2n5)=8(4n1)97(2n5)=8(4n1)9

276.

4(p4)(p+7)=5(p3)4(p4)(p+7)=5(p3)

277.

3(a2)(a+6)=4(a1)3(a2)(a+6)=4(a1)

278.

(9y+5)(3y7)(9y+5)(3y7)
=16(4y2)=16(4y2)

279.

(7m+4)(2m5)(7m+4)(2m5)
=14(5m3)=14(5m3)

280.

4[58(4c3)]4[58(4c3)]
=12(113c)8=12(113c)8

281.

5[92(6d1)]5[92(6d1)]
=11(410d)139=11(410d)139

282.

3[−9+8(4h3)]3[−9+8(4h3)]
=2(512h)19=2(512h)19

283.

3[−14+2(15k6)]3[−14+2(15k6)]
=8(35k)24=8(35k)24

284.

5[2(m+4)+8(m7)]5[2(m+4)+8(m7)]
=2[3(5+m)(213m)]=2[3(5+m)(213m)]

285.

10[5(n+1)+4(n1)]10[5(n+1)+4(n1)]
=11[7(5+n)(253n)]=11[7(5+n)(253n)]

286.

5(1.2u4.8)=−125(1.2u4.8)=−12

287.

4(2.5v0.6)=7.64(2.5v0.6)=7.6

288.

0.25(q6)=0.1(q+18)0.25(q6)=0.1(q+18)

289.

0.2(p6)=0.4(p+14)0.2(p6)=0.4(p+14)

290.

0.2(30n+50)=280.2(30n+50)=28

291.

0.5(16m+34)=−150.5(16m+34)=−15

Classify Equations

In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.

292.

23z+19=3(5z9)+8z+4623z+19=3(5z9)+8z+46

293.

15y+32=2(10y7)5y+4615y+32=2(10y7)5y+46

294.

5(b9)+4(3b+9)=6(4b5)7b+215(b9)+4(3b+9)=6(4b5)7b+21

295.

9(a4)+3(2a+5)=7(3a4)6a+79(a4)+3(2a+5)=7(3a4)6a+7

296.

18(5j1)+29=4718(5j1)+29=47

297.

24(3d4)+100=5224(3d4)+100=52

298.

22(3m4)=8(2m+9)22(3m4)=8(2m+9)

299.

30(2n1)=5(10n+8)30(2n1)=5(10n+8)

300.

7v+42=11(3v+8)2(13v1)7v+42=11(3v+8)2(13v1)

301.

18u51=9(4u+5)6(3u10)18u51=9(4u+5)6(3u10)

302.

3(6q9)+7(q+4)=5(6q+8)5(q+1)3(6q9)+7(q+4)=5(6q+8)5(q+1)

303.

5(p+4)+8(2p1)=9(3p5)6(p2)5(p+4)+8(2p1)=9(3p5)6(p2)

304.

12(6h1)=8(8h+5)412(6h1)=8(8h+5)4

305.

9(4k7)=11(3k+1)+49(4k7)=11(3k+1)+4

306.

45(3y2)=9(15y6)45(3y2)=9(15y6)

307.

60(2x1)=15(8x+5)60(2x1)=15(8x+5)

308.

16(6n+15)=48(2n+5)16(6n+15)=48(2n+5)

309.

36(4m+5)=12(12m+15)36(4m+5)=12(12m+15)

310.

9(14d+9)+4d=13(10d+6)+39(14d+9)+4d=13(10d+6)+3

311.

11(8c+5)8c=2(40c+25)+511(8c+5)8c=2(40c+25)+5

Everyday Math

312.

Fencing Micah has 44 feet of fencing to make a dog run in his yard. He wants the length to be 2.5 feet more than the width. Find the length, L, by solving the equation 2L+2(L2.5)=442L+2(L2.5)=44.

313.

Coins Rhonda has $1.90 in nickels and dimes. The number of dimes is one less than twice the number of nickels. Find the number of nickels, n, by solving the equation 0.05n+0.10(2n1)=1.900.05n+0.10(2n1)=1.90.

Writing Exercises

314.

Using your own words, list the steps in the general strategy for solving linear equations.

315.

Explain why you should simplify both sides of an equation as much as possible before collecting the variable terms to one side and the constant terms to the other side.

316.

What is the first step you take when solving the equation 37(y4)=3837(y4)=38 ? Why is this your first step?

317.

Solve the equation 14(8x+20)=3x414(8x+20)=3x4 explaining all the steps of your solution as in the examples in this section.

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objective of this section.

This is a table that has three rows and four columns. In the first row, which is a header row, the cells read from left to right: “I can…,” “confidently,” “with some help,” and “no-I don’t get it!” The first column below “I can…” reads: “solve equations using the general strategy for solving linear equations,” and “classify equations.” The rest of the cells are blank.

On a scale of 1-10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?

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