Section 2.1Use a General Strategy to Solve Linear Equations

Solve Linear Equations Using a General Strategy

Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that makes it a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!

To determine whether a number is a solution to an equation, we substitute the value for the variable in the equation. If the resulting equation is a true statement, then the number is a solution of the equation.

Example 1

Determine whether the values are solutions to the equation: \(5y+3=10y-4.\)

ⓐ \(y=\frac{3}{5}\) ⓑ \(y=\frac{7}{5}\)

Substitute each given value in for y and simplify both sides to see whether they match.

Show solution

Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.

ⓐ

Table 1

Multiply.
Simplify.

Since \(y=\frac{3}{5}\) does not result in a true equation, \(y=\frac{3}{5}\) is not a solution to the equation \(5y+3=10y-4.\)

ⓑ

Table 2

Multiply.
Simplify.

Since \(y=\frac{7}{5}\) results in a true equation, \(y=\frac{7}{5}\) is a solution to the equation \(5y+3=10y-4.\)

There are many types of equations that we will learn to solve. In this section we will focus on a linear equation.

To solve a linear equation it is a good idea to have an overall strategy that can be used to solve any linear equation. In the next example, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

Example 2How to Solve a Linear Equation Using a General Strategy

Solve: \(7(n-3)-8=-15\) .

Distribute the 7 across (n-3) to simplify the left side first.

Show solution

These steps are summarized in the General Strategy for Solving Linear Equations below.

Example 3

Solve: \(\frac{2}{3}(3m-6)=5-m.\)

Distribute the 2/3 across (3m-6) first to simplify the left side.

Show solution

Table 3

Distribute.
Add m to both sides to get the variables only on the left.
Simplify.
Add 4 to both sides to get constants only on the right.
Simplify.
Divide both sides by three.
Simplify.
Check:
Let \(m=3.\)

We can solve equations by getting all the variable terms to either side of the equal sign. By collecting the variable terms on the side where the coefficient of the variable is larger, we avoid working with some negatives. This will be a good strategy when we solve inequalities later in this chapter. It also helps us prevent errors with negatives.

Example 4

Solve: \(4(x-1)-2=5(2x+3)+6.\)

Distribute on both sides, then move the variable terms to the side with the larger coefficient.

Show solution

Table 4

Distribute.
Combine like terms.
Subtract \(4x\) from each side to get the variables only on the right since \(10>4.\)
Simplify.
Subtract 21 from each side to get the constants on left.
Simplify.
Divide both sides by 6.
Simplify.
Check:
Let \(x=-\frac{9}{2}.\)

Example 5

Solve: \(10[3-8(2s-5)]=15(40-5s).\)

Simplify inside the innermost parentheses first, then work outward.

Show solution

Table 5

Simplify from the innermost parentheses first.
Combine like terms in the brackets.
Distribute.
Add \(160s\) to both sides to get the variables to the right.
Simplify.
Subtract 600 from both sides to get the constants to the left.
Simplify.
Divide both sides by 85.
Simplify.
Check:
Let \(s=-2.\)

Classify Equations

Whether or not an equation is true depends on the value of the variable. The equation \(7x+8=-13\) is true when we replace the variable, x, with the value \(-3,\) but not true when we replace x with any other value. An equation like this is called a conditional equation. All the equations we have solved so far are conditional equations.

Now let’s consider the equation \(7y+14=7(y+2).\) Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.

Solve:

Table 6

Distribute.
Subtract \(7y\) to each side to get the \(y’\text{s}\) to one side.
Simplify—the y’s are eliminated.
	But \(14=14\) is true.

This means that the equation \(7y+14=7(y+2)\) 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 is called an identity.

What happens when we solve the equation \(-8z=-8z+9?\)

Solve:

Table 7

Add \(8z\) to both sides to leave the constant alone on the right.
Simplify—the \(z’\text{s}\) are eliminated.
	But \(0\ne 9.\)

Solving the equation \(-8z=-8z+9\) led to the false statement \(0=9.\) The equation \(-8z=-8z+9\) 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.

The next few examples will ask us to classify an equation as conditional, an identity, or as a contradiction.

Example 6

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(6(2n-1)+3=2n-8+5(2n+1).\)

Distribute and combine like terms on each side to see what happens to the variable.

Show solution

Table 8

Distribute.
Combine like terms.
Subtract \(12n\) from each side to get the n’s to one side.
Simplify.
This is a true statement.	The equation is an identity.
	The solution is all real numbers.

Example 7

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(8+3(a-4)=0.\)

Distribute and combine like terms, then solve for a as usual.

Show solution

Table 9

Distribute.
Combine like terms.
Add 4 to both sides.
Simplify.
Divide.
Simplify.
The equation is true when \(a=\frac{4}{3}.\,\)	This is a conditional equation.
	The solution is \(a=\frac{4}{3}.\)

Example 8

Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(5m+3(9+3m)=2(7m-11).\)

Distribute and combine like terms on each side before comparing them.

Show solution

Table 10

Distribute.
Combine like terms.
Subtract \(14m\) from both sides.
Simplify.
But \(27\ne \text{-}22.\)	The equation is a contradiction.
	It has no solution.

We summarize the methods for classifying equations in the table.

Solve Equations with Fraction or Decimal Coefficients

We could use the General Strategy to solve the next example. This method would work fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.

We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator (LCD) of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but without fractions. This process is called clearing the equation of fractions.

To clear an equation of decimals, we think of all the decimals in their fraction form and then find the LCD of those denominators.

Example 9How to Solve Equations with Fraction or Decimal Coefficients

Solve: \(\frac{1}{12}x+\frac{5}{6}=\frac{3}{4}.\)

Find the LCD of all the fractions, then multiply both sides by it to clear them.

Show solution

Notice in the previous example, once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve. We then used the General Strategy for Solving Linear Equations.

Example 10

Solve: \(5=\frac{1}{2}y+\frac{2}{3}y-\frac{3}{4}y.\)

Find the LCD of the three fractions, then multiply every term by it.

Show solution

We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.

Table 12

Find the LCD of all fractions in the equation.
The LCD is 12.
Multiply both sides of the equation by 12.
Distribute.
Simplify—notice, no more fractions.
Combine like terms.
Divide by five.
Simplify.
Check:
Let \(y=12.\)

In the next example, we’ll distribute before we clear the fractions.

Example 11

Solve: \(\frac{1}{2}(y-5)=\frac{1}{4}(y-1).\)

Multiply both sides by the LCD to clear the fractions before distributing.

Show solution

Table 13

Distribute.
Simplify.
Multiply by the LCD, four.
Distribute.
Simplify.
Collect the variables to the left.
Simplify.
Collect the constants to the right.
Simplify.
An alternate way to solve this equation is to clear the fractions without distributing first. If you multiply the factors correctly, this method will be easier.
Multiply by the LCD, 4.
Multiply four times the fractions.
Distribute.
Collect the variables to the left.
Simplify.
Collect the constants to the right.
Simplify.
Check:
Let \(y=9.\)
Finish the check on your own.

When you multiply both sides of an equation by the LCD of the fractions, make sure you multiply each term by the LCD—even if it does not contain a fraction.

Example 12

Solve: \(\frac{4q+3}{2}+6=\frac{3q+5}{4}\)

Multiply every term, including the 6, by the LCD to clear the fractions.

Show solution

Table 14

Multiply both sides by the LCD, 4.
Distribute.
Simplify.
Collect the variables to the left.
Simplify.
Collect the constants to the right.
Simplify.
Divide both sides by five.
Simplify.
Check:
Let \(q=-5.\)
Finish the check on your own.

Some equations have decimals in them. This kind of equation may occur when we solve problems dealing with money or percentages. But decimals can also be expressed as fractions. For example, \(0.7=\frac{7}{10}\) and \(0.29=\frac{29}{100}.\) So, with an equation with decimals, we can use the same method we used to clear fractions—multiply both sides of the equation by the least common denominator.

The next example uses an equation that is typical of the ones we will see in the money applications in a later section. Notice that we will clear all decimals by multiplying by the LCD of their fraction form.

Example 13

Solve: \(0.25x+0.05(x+3)=2.85.\)

Rewrite each decimal as a fraction to find the LCD, then multiply both sides by it.

Show solution

Look at the decimals and think of the equivalent fractions:

\[0.25=\frac{25}{100},\,0.05=\frac{5}{100},\,2.85=2\frac{85}{100}.\]

Notice, the LCD is 100. By multiplying by the LCD we will clear the decimals from the equation.

Table 15

Distribute first.
Combine like terms.
To clear decimals, multiply by 100.
Distribute.
Subtract 15 from both sides.
Simplify.
Divide by 30.
Simplify.
Check it yourself by substituting \(x=9\) into the original equation.

Key Concepts

• How to determine whether a number is a solution to an equation
  • Substitute the number in for the variable in the equation.
  • Simplify the expressions on both sides of the equation.
  • Determine whether the resulting equation is true.
    If it is true, the number is a solution.
    If it is not true, the number is not a solution.
  See Example 1.
• How to Solve Linear Equations Using a General Strategy
  • Simplify each side of the equation as much as possible.
    Use the Distributive Property to remove any parentheses.
    Combine like terms.
  • Collect all the variable terms on one side of the equation.
    Use the Addition or Subtraction Property of Equality.
  • Collect all the constant terms on the other side of the equation.
    Use the Addition or Subtraction Property of Equality.
  • Make the coefficient of the variable term equal to 1.
    Use the Multiplication or Division Property of Equality.
    State the solution to the equation.
  • Check the solution.
    Substitute the solution into the original equation to make sure the result is a true statement.
  See Example 2, Example 3, Example 4, and Example 5.
• How to Solve Equations with Fraction or Decimal Coefficients
  • Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
  • Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
  • Solve using the General Strategy for Solving Linear Equations.
  See Example 9, Example 10, Example 11, Example 12, and Example 13.

Section Exercises

Practice Makes Perfect

Solve Equations Using the General Strategy

In the following exercises, determine whether the given values are solutions to the equation.

4

\(6y+10=12y\)

ⓐ \(y=\frac{5}{3}\) ⓑ \(y=-\frac{1}{2}\)

Show answer

ⓐ yes ⓑ no

5

\(4x+9=8x\)

ⓐ \(x=-\frac{7}{8}\) ⓑ \(x=\frac{9}{4}\)

6

\(8u-1=6u\)

ⓐ \(u=-\frac{1}{2}\) ⓑ \(u=\frac{1}{2}\)

Show answer

ⓐ no ⓑ yes

7

\(9v-2=3v\)

ⓐ \(v=-\frac{1}{3}\) ⓑ \(v=\frac{1}{3}\)

In the following exercises, solve each linear equation.

8

\(15(y-9)=-60\)

Show answer

\(y=5\)

9

\(-16(3n+4)=32\)

10

\(\text{-}(w-12)=30\)

Show answer

\(w=-18\)

11

\(\text{-}(t-19)=28\)

12

\(51+5(4-q)=56\)

Show answer

\(q=3\)

13

\(-6+6(5-k)=15\)

14

\(3(10-2x)+54=0\)

Show answer

\(x=14\)

15

\(-2(11-7x)+54=4\)

16

\(\frac{2}{3}(9c-3)=22\)

Show answer

\(c=4\)

17

\(\frac{3}{5}(10x-5)=27\)

18

\(\frac{1}{5}(15c+10)=c+7\)

Show answer

\(c=\frac{5}{2}\)

19

\(\frac{1}{4}(20d+12)=d+7\)

20

\(3(4n-1)-2=8n+3\)

Show answer

\(n=2\)

21

\(9(2m-3)-8=4m+7\)

22

\(12+2(5-3y)=-9(y-1)-2\)

Show answer

\(y=-5\)

23

\(-15+4(2-5y)=-7(y-4)+4\)

24

\(5+6(3s-5)=-3+2(8s-1)\)

Show answer

\(s=10\)

25

\(-12+8(x-5)=-4+3(5x-2)\)

26

\(4(p-4)-(p+7)=5(p-3)\)

Show answer

\(p=-4\)

27

\(3(a-2)-(a+6)=4(a-1)\)

28

\(4[5-8(4c-3)]=12(1-13c)-8\)

Show answer

\(c=-4\)

29

\(5[9-2(6d-1)]=11(4-10d)-139\)

30

\(3[-9+8(4h-3)]=2(5-12h)-19\)

Show answer

\(h=\frac{3}{4}\)

31

\(3[-14+2(15k-6)]=8(3-5k)-24\)

32

\(5[2(m+4)+8(m-7)]=2[3(5+m)-(21-3m)]\)

Show answer

\(m=6\)

33

\(10[5(n+1)+4(n-1)]=11[7(5+n)-(25-3n)]\)

Classify Equations

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

34

\(23z+19=3(5z-9)+8z+46\)

Show answer

identity; all real numbers

35

\(15y+32=2(10y-7)-5y+46\)

36

\(18(5j-1)+29=47\)

Show answer

conditional equation; \(j=\frac{2}{5}\)

37

\(24(3d-4)+100=52\)

38

\(22(3m-4)=8(2m+9)\)

Show answer

conditional equation; \(m=\frac{16}{5}\)

39

\(30(2n-1)=5(10n+8)\)

40

\(7v+42=11(3v+8)-2(13v-1)\)

Show answer

contradiction; no solution

41

\(18u-51=9(4u+5)-6(3u-10)\)

42

\(45(3y-2)=9(15y-6)\)

Show answer

contradiction; no solution

43

\(60(2x-1)=15(8x+5)\)

44

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

Show answer

identity; all real numbers

45

\(11(8c+5)-8c=2(40c+25)+5\)

Solve Equations with Fraction or Decimal Coefficients

In the following exercises, solve each equation with fraction coefficients.

46

\(\frac{1}{4}x-\frac{1}{2}=-\frac{3}{4}\)

Show answer

\(x=-1\)

47

\(\frac{3}{4}x-\frac{1}{2}=\frac{1}{4}\)

48

\(\frac{5}{6}y-\frac{2}{3}=-\frac{3}{2}\)

Show answer

\(y=-1\)

49

\(\frac{5}{6}y-\frac{1}{3}=-\frac{7}{6}\)

50

\(\frac{1}{2}a+\frac{3}{8}=\frac{3}{4}\)

Show answer

\(a=\frac{3}{4}\)

51

\(\frac{5}{8}b+\frac{1}{2}=-\frac{3}{4}\)

52

\(2=\frac{1}{3}x-\frac{1}{2}x+\frac{2}{3}x\)

Show answer

\(x=4\)

53

\(2=\frac{3}{5}x-\frac{1}{3}x+\frac{2}{5}x\)

54

\(\frac{1}{3}w+\frac{5}{4}=w-\frac{1}{4}\)

Show answer

\(w=\frac{9}{4}\)

55

\(\frac{1}{2}a-\frac{1}{4}=\frac{1}{6}a+\frac{1}{12}\)

56

\(\frac{1}{3}b+\frac{1}{5}=\frac{2}{5}b-\frac{3}{5}\)

Show answer

\(b=12\)

57

\(\frac{1}{3}x+\frac{2}{5}=\frac{1}{5}x-\frac{2}{5}\)

58

\(\frac{1}{4}(p-7)=\frac{1}{3}(p+5)\)

Show answer

\(p=-41\)

59

\(\frac{1}{5}(q+3)=\frac{1}{2}(q-3)\)

60

\(\frac{1}{2}(x+4)=\frac{3}{4}\)

Show answer

\(x=-\frac{5}{2}\)

61

\(\frac{1}{3}(x+5)=\frac{5}{6}\)

62

\(\frac{4n+8}{4}=\frac{n}{3}\)

Show answer

\(n=-3\)

63

\(\frac{3p+6}{3}=\frac{p}{2}\)

64

\(\frac{3x+4}{2}+1=\frac{5x+10}{8}\)

Show answer

\(x=-2\)

65

\(\frac{10y-2}{3}+3=\frac{10y+1}{9}\)

66

\(\frac{7u-1}{4}-1=\frac{4u+8}{5}\)

Show answer

\(u=3\)

67

\(\frac{3v-6}{2}+5=\frac{11v-4}{5}\)

In the following exercises, solve each equation with decimal coefficients.

68

\(0.4x+0.6=0.5x-1.2\)

Show answer

\(x=18\)

69

\(0.7x+0.4=0.6x+2.4\)

70

\(0.9x-1.25=0.75x+1.75\)

Show answer

\(x=20\)

71

\(1.2x-0.91=0.8x+2.29\)

72

\(0.05n+0.10(n+8)=2.15\)

Show answer

\(n=9\)

73

\(0.05n+0.10(n+7)=3.55\)

74

\(0.10d+0.25(d+5)=4.05\)

Show answer

\(d=8\)

75

\(0.10d+0.25(d+7)=5.25\)

Everyday Math

76

Fencing Micah has 74 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 \(2L+2(L-2.5)=74.\)

Show answer

\(L=19.75\) feet

77

Stamps Paula bought $22.82 worth of 49-cent stamps and 21-cent stamps. The number of 21-cent stamps was eight less than the number of
49-cent stamps. Solve the equation
\(0.49s+0.21\,(s-8)\,\,=22.82\) for s, to find the number of 49-cent stamps Paula bought.

Writing Exercises

78

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

Show answer

Answers will vary.

79

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.

80

What is the first step you take when solving the equation \(3-7(y-4)=38?\) Why is this your first step?

Show answer

Answers will vary.

81

If an equation has several fractions, how does multiplying both sides by the LCD make it easier to solve?

82

If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?

Show answer

Answers will vary.

83

For the equation \(0.35x+2.1=3.85,\) how do you clear the decimal?

Self Check

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

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

Glossary

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.

contradiction

An equation that is false for all values of the variable is called a contradiction. A contradiction has no solution.

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.

linear equation

A linear equation is an equation in one variable that can be written, where a and b are real numbers and \(a\ne 0,\) as \(ax+b=0.\)

solution of an equation

A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.