2.4 Use a General Strategy to Solve Linear Equations

Learning objectives

Learning objectives

• Solve equations using a general strategy

• Classify equations

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

Example 2.38

Solution

Example 2.39

Solution

Example 2.40

Solution

Example 2.41

Solution

Example 2.42

Solution

Example 2.43

Solution

Example 2.44

Solution

Classify Equations

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

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

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. Whether or not the equation $7x+8=-13$ 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.

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

But $6=6$ is true.

This means that the equation $2y+6=2(y+3)$ 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.

What happens when we solve the equation $5z=5z-1$?

But $0\ne-1$.

Solving the equation $5z=5z-1$ led to the false statement $0=-1$. The equation $5z=5z-1$ 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.

Example 2.45

Solution

Example 2.46

Solution

Example 2.47

Solution

Table

2.5

Section 2.4 Exercises

Section 2.4 Exercises

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, solve each linear equation.

Classify Equations

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

Everyday Math

Writing Exercises

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

ⓑ 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?

1. Solve: $5(x+3)=35.$

2. Solve: $6(y-4)=-18.$

3. Solve: $-(y+8)=-2.$

4. Solve: $-(z+4)=-12.$

5. Solve: $2(m-4)+3=-1$.

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

2. Solve: $\frac{1}{3}(6u+3)=7-u$.

3. Solve: $\frac{2}{3}(9x-12)=8+2x$.

4. Solve: $12-3(4j+3)=-17$.

5. Solve: $-6-8(k-2)=-10$.

1. Solve: $6(p-3)-7=5(4p+3)-12$.

2. Solve: $8(q+1)-5=3(2q-4)-1$.

3. Solve: $6[4-2(7y-1)]=8(13-8y)$.

4. Solve: $12[1-5(4z-1)]=3(24+11z)$.

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

1. $-(t-19)=28$

2. $8(9b-4)-12=100$

3. $21+2(m-4)=25$

4. $-6+6(5-k)=15$

5. $8(6t-5)-35=-27$

1. $-2(11-7x)+54=4$

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

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

4. $15-(3r+8)=28$

5. $-3-(m-1)=13$

1. $18-2(y-3)=32$

2. $35-5(2w+8)=-10$

3. $-2(a-6)=4(a-3)$

4. $5(8-r)=-2(2r-16)$

5. $9(2m-3)-8=4m+7$

1. $-15+4(2-5y)=-7(y-4)+4$

2. $5(x-4)-4x=14$

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

4. $7(2n-5)=8(4n-1)-9$

5. $3(a-2)-(a+6)=4(a-1)$

1. $-(7m+4)-(2m-5)$$=14-(5m-3)$

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

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

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

5. $4(2.5v-0.6)=7.6$

1. $0.2(p-6)=0.4(p+14)$

2. $0.5(16m+34)=-15$

3. $15y+32=2(10y-7)-5y+46$

4. $9(a-4)+3(2a+5)=7(3a-4)-6a+7$

5. $24(3d-4)+100=52$

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

2. $18u-51=9(4u+5)-6(3u-10)$

3. $5(p+4)+8(2p-1)=9(3p-5)-6(p-2)$

4. $9(4k-7)=11(3k+1)+4$

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

1. Solve: $0.15(40m-120)=0.5(60m+12)$.

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

$4+9(3x-7)=-42x-13+23(3x-2)$

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

$8(1-3x)+15(2x+7)=2(x+50)+4(x+3)+1$

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

5. Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: $6+14(k-8)=95$

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

$12c+5(5+3c)=3(9c-4)$

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

$4(7d+18)=13(3d-2)-11d$

3. $21(y-5)=-42$

4. $-16(3n+4)=32$

5. $5(8+6p)=0$

1. $36(4m+5)=12(12m+15)$

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

3. 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$0.05n+0.10(2n-1)=1.90$.

4. 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.

5. Solve the equation $\frac{1}{4}(8x+20)=3x-4$ explaining all the steps of your solution as in the examples in this section.