# 3.4 Use a General Strategy to Solve Linear Equations

## Learning Objectives

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

• Solve equations using a general strategy
• Classify equations

## Try It

Before you get started, take this readiness quiz:

1) Simplify: $−(a-4)$
2) Multiply: $\frac{3}{2}(12x+20)$.
3) Simplify: $5-2(n+1)$.
4) Multiply: $3(7y+9)$.
5) Multiply: $(2.5)(6.4)$.

## 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 3.4.1

Solve: $-6(x+3)=24$.

Solution

Step 1: Simplify each side of the equation as much as possible.

Use the Distributive Property.

\begin{align*} -6\left(x+3\right)&=24\\ -6x-18&=24 \end{align*}

Notice that each side of the equation is simplified as much as possible.

Step 2: Collect all variable terms on one side of the equation.

Nothing to to – all $x$‘s are on the left side.

Step 3: Collect constant terms on the other side of the equation.

To get constants only on the right, add 18 to each side.

\begin{align*}&\;&-6x-18{\color{red}{+}}{\color{red}{18}}&=24{\color{red}{+}}{\color{red}{18}}\\ &\text{Simplify.}\;&-6x&=42 \end{align*}

Step 4: Make the coefficient of the variable term to equal 1.

\begin{align*} &\text{Divide each side by -6.}\;&\frac{-6x}{{\color{red}{-}}{\color{red}{6}}}&=\frac{42}{{\color{red}{-}}{\color{red}{6}}}\\ &\text{Simplify.}\;&x&=-7 \end{align*}

Step 5: Check the solution.

Let $x=-7$

$\begin{eqnarray*}-6\left(x+3\right)=\;&24&\\-6\left({\color{red}{-}}{\color{red}{7}}+3\right)\overset?=\;&24&\\\text{Simplify.}-6\left(-4\right)\overset?=\;&24&\\\text{Multiply.}\;24=\;&24&\end{eqnarray*}$

### Try It

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

Solution

$x=4$

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

Solution

$y=1$

### HOW TO

#### General strategy for solving linear equations.

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

### Example 3.4.2

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

Solution

Step 1: Simplify each side of the equation as much as possible by distributing.

$-y-9=8$

Step 2: The only y term is on the left side, so all variable terms are on the left side of the equation.

Step 3: Add 9 to both sides to get all constant terms on the right side of the equation.

\begin{align*}&\;&-y-9{\color{red}{+}}{\color{red}{9}}&=8{\color{red}{+}}{\color{red}{9}}\\ &\text{Simplify.}\;&-y&=17 \end{align*}

Step 4: Rewrite $-y$ as $-1y$.

$-1y=17$

Step 5: Make the coefficient of the variable term to equal to 1 by dividing both sides by -1.

\begin{align*}&\;&\frac{-1y}{-1}&=\frac{17}{-1}\\ &\text{Simplify.}\;&y&=-17 \end{align*}

Step 6: Check:

$-(y+9)=8$

Step 7: Let $y=-17$.

\begin{align*} -\left({\color{red}{-}}{\color{red}{17}}+9\right)&\overset?=8\\ -\left(-8\right)&\overset?=8\\ 8&=8\checkmark \end{align*}

### Try It

8) Solve: $−(y+8)=-2$.

Solution

$y=-6$

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

Solution

$z=8$

### Example 3.4.3

Solve: $5(a-3)+5=-10$.

Solution

Step 1: Simplify each side of the equation as much as possible.

\begin{align*} &\text{Distribute.}\;&5a-15+5&=-10\\ &\text{Combine like terms.}\;&5a-10&=-10 \end{align*}

Step 2: The only $a$ term is on the left side, so all variable terms are on one side of the equation.

Step 3: Add 10 to both sides to get all constant terms on the other side of the equation.

\begin{align*} &\;&5a-10{\color{red}{+}}{\color{red}{10}}&=-10{\color{red}{+}}{\color{red}{10}}\\ &\text{Simplify.}\;&5a&=0\\ \end{align*}

Step 4: Make the coefficient of the variable term to equal to 1 by dividing both sides by 5.

\begin{align*} &\;&\frac{5a}{\color{red}{5}}&=\frac0{\color{red}{5}}\\ &\text{Simplify.}\;&a&=0\\ \end{align*}

Step 5: Check:

$5(a-3)+5=-10$

Step 6: Let $a=0$.

\begin{align*} 5\left({\color{red}{0}}-3\right)+5&\overset?=-10\\ 5\left(-3\right)+5&\overset?=-10\\ -15+5&\overset?=-10\\ -10&=-10\checkmark \end{align*}

### Try It

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

Solution

$m=2$

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

Solution

$n=2$

### Example 3.4.4

Solve: $\frac{2}{3}(6m-3)=8-m$.

Solution

Step 1: Distribute.

$4m-2=8-m$

Step 2: Add $m$ to get the variables only to the left.

\begin{align*} &\;&4m{\color{red}{+}}{\color{red}{m}}-2&=8-m{\color{red}{+}}{\color{red}{m}}\\ &\text{Simplify.}\;&5m-2&=8\\ \end{align*}

Step 3: Add 2 to get constants only on the right.

\begin{align*} &\;&5m-2{\color{red}{+}}{\color{red}{2}}&=8{\color{red}{+}}{\color{red}{2}}\\ &\text{Simplify.}\;&5m&=10\\ \end{align*}

Step 4: Divide by 5.

\begin{align*} &\;&\frac{5m}{\color{red}{5}}&=\frac{10}{\color{red}{5}}\\ &\text{Simplify.}\;&m&=2\\ \end{align*}

Step 5: Check:

$\frac23(6m-3)=8-m$

Step 6: Let $m=2$.

\begin{align*} \frac23\left(6\cdot{\color{red}{2}}\cdot-3\right)&\overset?=8-{\color{red}{2}}\\ \frac23\left(12-3\right)&\overset?=6\\ \frac23\left(9\right)&\overset?=6\\ 6&=6\checkmark \end{align*}

### Try It

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

Solution

$u=2$

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

Solution

$x=4$

### Example 3.4.5

Solve: $8-2(3y+5)=0$.
Solution

Step 1: Simplify—use the Distributive Property.

\begin{align*} &\;&8-6y-10&=0\\ &\text{Combine like terms.}\;&-6y-2&=0\\ \end{align*}

Step 2: Add 2 to both sides to collect constants on the right.

\begin{align*} &\;&-6y-2{\color{red}{+}}{\color{red}{2}}&=0{\color{red}{+}}{\color{red}{2}}\\ &\text{Simplify.}\;&-6y&=2\\ \end{align*}

Step 3: Divide both sides by $-6$.

\begin{align*} &\;&\frac{-6y}{{\color{red}{-}}{\color{red}{6}}}&=\frac2{{\color{red}{-}}{\color{red}{6}}}\\ &\text{Simplify.}\;&y&=-\frac13\\ \end{align*}

Step 4: Check:

Let $y=-\frac{1}{3}$

\begin{align*} 8-2\left(3y+5\right)&=0\\ 8-2\left[3\left({\color{red}{-}}{\color{red}{\frac13}}\right)+5\right]&=0\\ 8-2\left(-1+5\right)&\overset?=0\\ 8-2\left(4\right)&\overset?=0\\ 8-8&\overset?=0\\ 0&=0\checkmark \end{align*}

### Try It

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

Solution

$j=\frac{5}{3}$

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

Solution

$k=\frac{5}{2}$

### Example 3.4.6

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

Solution

Step 1: Distribute.

\begin{align*} &\;&4x-4-2&=10x+15+6\\ &\text{Combine like terms.}\;&4x-6&=10x+21\\ \end{align*}

Step 2: Subtract $4x$ to get the variables only on the right side since $10>4$.

\begin{align*} &\;&4x{\color{red}{-}}{\color{red}{4}}{\color{red}{x}}-6&=10{\color{red}{-}}{\color{red}{4}}{\color{red}{x}}+21\\ &\text{Simplify.}\;&-6&=6x+21\\ \end{align*}

Step 3: Subtract 21 to get the constants on left.

\begin{align*} &\;&6{\color{red}{-}}{\color{red}{21}}&=6x+21{\color{red}{-}}{\color{red}{21}}\\ &\text{Simplify.}\;&-27&=6x\\ \end{align*}

Step 4: Divide by 6.

\begin{align*} &\;&\frac{-27}{\color{red}{6}}&=\frac{6x}{\color{red}{6}}\\ &\text{Simplify.}\;&-\frac92&=x\\ \end{align*}

Step 5: Check:

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

Step 6: Let $x=-\frac{9}{2}$.

\begin{align*} 4\left({\color{red}{-}}{\color{red}{\frac92}}\right)-2&\overset?=5\left[2\left({\color{red}{-}}{\color{red}{\frac92}}\right)+3\right]+6\\ 4\left(-\frac{11}2\right)-2&\overset?=5\left(-9+3\right)+6\\ -22-2&\overset?=5\left(-6\right)+6\\ -24&\overset?=-30+6\\ -24&=-24\checkmark \end{align*}

### Try It

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

Solution

$p=-2$

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

Solution

$q=-8$

### Example 3.4.7

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

Solution

Step 1: Simplify from the innermost parentheses first.

$10[3-16s+40]=15(40-5s)$

Step 2: Combine like terms in the brackets.

$10[43-16s]=15(40-5s)$

Step 3: Distribute.

$430-160s=600-75s$

Step 4: Add $160s$ to get the $s$’s to the right.

\begin{align*} &\;&430-160{\color{red}{+}}{\color{red}{160}}{\color{red}{s}}&=600-85s{\color{red}{+}}{\color{red}{160}}{\color{red}{s}}\\ &\text{Simplify.}\;&430&=600+85s\\ \end{align*}

Step 5: Subtract 600 to get the constants to the left.

\begin{align*} &\;&430-{\color{red}{600}}&=85s+600{\color{red}{-}}{\color{red}{600}}\\ &\text{Simplify.}\;&-170&=85s\\ \end{align*}

Step 6: Divide.

\begin{align*} &\;&\frac{-170}{\color{red}{85}}&=\frac{85s}{\color{red}{85}}\\ &\text{Simplify.}\;&-2&=s\\ \end{align*}

Step 7: Check:

$10[3-8(2s-5)]=15(40-5s)$

Step 8: Substitute $s=-2$.

\begin{align*} 10\left[3-8\left(2{\color{black}{\left({\color{red}{-}}{\color{red}{2}}\right)}}-5\right)\right]&\overset?=15\left(40-5\left({\color{red}{-}}{\color{red}{2}}\right)\right)\\ 10\left[3-8\left(-4-5\right)\right]&\overset?=15\left(40+10\right)\\ 10\left[3-8\left(-9\right)\right]&\overset?=15\left(50\right)\\ 10\left[3+72\right]&\overset?=750\\ 10\left[75\right]&\overset?=750\\ 750&=750\checkmark \end{align*}

### Try It

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

Solution

$y=-\frac{17}{5}$

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

Solution

$z=0$

### Example 3.4.8

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

Solution

Step 1: Distribute.

$36n+1.8=18n+9$

Step 2: Subtract $18n$ to get the variables to the left.

\begin{align*} &\;&36n{\color{red}{-}}{\color{red}{18}}{\color{red}{n}}+1.8&=18n{\color{red}{-}}{\color{red}{18}}{\color{red}{n}}+9\\ &\text{Simplify.}\;&18n+1.8&=9\\ \end{align*}

Step 3: Subtract 1.8 to get the constants to the right.

\begin{align*} &\;&18n+1.8{\color{red}{-}}{\color{red}{1}}{\color{red}{.}}{\color{red}{8}}&=9{\color{red}{-}}{\color{red}{1}}{\color{red}{.}}{\color{red}{8}}\\ &\text{Simplify.}\;&18n&=7.2\\ \end{align*}

Step 4: Divide.

\begin{align*} &\;&\frac{18n}{\color{red}{18}}&=\frac{7.2}{\color{red}{18}}\\ &\text{Simplify.}\;&n&=0.4\\ \end{align*}

Step 5: Check:

$0.35(100n+5)=0.6(30n+15)$

Step 6: Let $n=0.4$.

\begin{align*} 0.36\left(100\left({\color{red}{0}}{\color{red}{.}}{\color{red}{4}}\right)+5\right)&\overset?=0.6\left(30\left({\color{red}{0}}{\color{red}{.}}{\color{red}{4}}\right)+15\right)\\ 0.36\left(40+5\right)&\overset?=0.6\left(12+15\right)\\ 0.36\left(45\right)&\overset?=0.6\left(27\right)\\ 16.2&=16.2\checkmark \end{align*}

### Try It

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

Solution

$n=1$

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

Solution

$m=-1$

## 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 $x=-3$.

\begin{align*} 7\left({\color{red}{-}}{\color{red}{3}}\right)+8&\overset?=-13\\ -21+8&\overset?=-13\\ -13&=-13\checkmark \end{align*}

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.

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

\begin{alignat}{3} &\;&\;\;\;\;\;2y+6&=2(y+3)\\ &\text{Distribute}&\;\;\;\;\;-2y+6&=2y+6\\ &\text{Subtract}\;2y\;\text{to get the}\;y's\;\text{to one side.}&\;\;\;\;\;2y{\color{red}{-}}{\color{red}{2}}{\color{red}{y}}+6&=2{\color{red}{-}}{\color{red}{2}}{\color{red}{y}}+6\\ &\text{Simplify-the}\;y's\;\text{are gone!}&\;\;\;\;\;6&=6 \end{alignat}

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 a 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 $5z=5z-1$?

\begin{alignat}{3} &\;&\;\;\;\;\;5z&=5z-1\\ &\text{Subtract}\;5z\;\text{to get the constant alone on the right.}&\;\;\;\;\;5z{\color{red}{-}}{\color{red}{5}}{\color{red}{z}}&=5z{\color{red}{-}}{\color{red}{5}}{\color{red}{z}}-1\\ &\text{Simplify-the}\;z's\;\text{are gone!}&\;\;\;\;\;0&\neq-1 \end{alignat}

But $0\neq{−1}$

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.

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

### Example 3.4.9

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

$6(2n-1)+3=2n-8+5(2n+1)$

Solution

Step 1: Distribute.

\begin{align*} &\;&12n-6+3&=2n-8+10n+5\\ &\text{Combine like terms.}\;&12n-3&=12n-3\\ \end{align*}

Step 2: Subtract $12n$ to get the $n$’s to one side.

\begin{align*} &\;&12n{\color{red}{-}}{\color{red}{12}}{\color{red}{n}}-3&=12n{\color{red}{-}}{\color{red}{12}}{\color{red}{n}}-3\\ &\text{Simplify.}\;&-3&=-3\\ \end{align*}

This is a true statement.

The equation is an identity.
The solution is all real numbers.

### Try It

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

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

Solution

identity; all real numbers

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

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

Solution

identity; all real numbers

### Example 3.4.10

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

$10+4(p-5)=0$

Solution

Step 1: Distribute.

\begin{align*} &\;&10+4p-20&=0\\ &\text{Combine like terms.}\;&4p-10&=0\\ \end{align*}

Step 2: Add 10 to both sides.

\begin{align*} &\;&4p-10{\color{red}{+}}{\color{red}{10}}&=0{\color{red}{+}}{\color{red}{10}}\\ &\text{Simplify.}\;&4p&=10\\ \end{align*}

Step 3: Divide.

\begin{align*} &\;&\frac{4p}{\color{red}{4}}&=\frac{10}{\color{red}{4}}\\ &\text{Simplify.}\;&p&=\frac52\\ \end{align*}

The equation is true when $p=\frac{5}{2}$.

This is a conditional equation.
The solution is $p=\frac{5}{2}$.

### Try It

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

Solution

conditional equation; $q=\frac{9}{11}$

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

Solution

conditional equation; $k=\frac{193}{14}$

### Example 3.4.11

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

$5m+3(9+3m)=2(7m-11)$

Solution

Step 1: Distribute.

\begin{align*} &\;&5m+27+9m&=14m-22\\ &\text{Combine like terms.}\;&14m+27&=14m-22\\ \end{align*}

Step 2: Subtract $14m$ from both sides.

\begin{align*} &\;&14m+27{\color{red}{-}}{\color{red}{14}}{\color{red}{m}}&=14m-22{\color{red}{-}}{\color{red}{14}}{\color{red}{m}}\\ &\text{Simplify.}\;&27&\neq-22\\ \end{align*}

But $27\neq-22.$

It has no solution.

### Try It

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

$5m+3(9+3m)=2(7m-11)$

Solution

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

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

Solution

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

### Key Concepts

• General Strategy for Solving Linear Equations
1. Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms.
2. Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality.
3. Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality.
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. Check the solution.
Substitute the solution into the original equation.

### Self Check

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?

### 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.