8.3 Solve Equations with Variables and Constants on Both Sides

Learning objectives

Learning objectives

• Solve an equation with constants on both sides

• Solve an equation with variables on both sides

• Solve an equation with variables and constants on both sides

• Solve equations using a general strategy

Solve an Equation with Constants on Both Sides

You may have noticed that in all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we’ll see how to solve equations where the variable terms and/or constant terms are on both sides of the equation.

Our strategy will involve choosing one side of the equation to be the variable side, and the other side of the equation to be the constant side. Then, we will use the Subtraction and Addition Properties of Equality, step by step, to get all the variable terms together on one side of the equation and the constant terms together on the other side.

By doing this, we will transform the equation that started with variables and constants on both sides into the form $ax=b.$ We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.

Example 8.20

Solution

1. Identify the constant attached to the variable term. In $4x+6=-14$, the number $6$ is added to $4x$, so subtracting $6$ from both sides will isolate the variable term.
2. Apply the Subtraction Property of Equality. Subtract $6$ from both sides: $4x+6-6=-14-6$, which preserves equality and cancels the added constant.
3. Apply the Division Property of Equality. The equation becomes $4x=-20$. Divide both sides by $4$: $\frac{4x}{4}=\frac{-20}{4}$, so $x=-5$.
4. Check by substitution and state the result. Substitute $x=-5$: $4(-5)+6=-20+6=-14$, so the solution is $x=-5$.

Takeaway: When a variable appears on only one side of an equation, undo addition or subtraction on both sides; here $x=-5$.

Example 8.21

Solution

1. Identify the operation attached to the variable term. In $2y-7=15$, the number $7$ is subtracted from $2y$, so adding $7$ to both sides will undo that subtraction.
2. Apply the Addition Property of Equality. Add $7$ to both sides: $2y-7+7=15+7$, which preserves equality while canceling the negative constant on the left.
3. Apply the Division Property of Equality. The equation becomes $2y=22$. Divide both sides by $2$: $\frac{2y}{2}=\frac{22}{2}$, so $y=11$.
4. Check by substitution and state the result. Substitute $y=11$: $2(11)-7=22-7=15$, so the solution is $y=11$.

Takeaway: Adding $7$ to both sides undoes the subtraction and gives $2y=22$, so $y=11$.

Solve an Equation with Variables on Both Sides

What if there are variables on both sides of the equation? We will start like we did above—choosing a variable side and a constant side, and then use the Subtraction and Addition Properties of Equality to collect all variables on one side and all constants on the other side. Remember, what you do to the left side of the equation, you must do to the right side too.

Example 8.22

Solution

1. Cancel the variable terms. Subtract $4x$ from both sides of $5x=4x+7$: $5x-4x=4x-4x+7$, which gives $x=7$.
2. Check by substitution. Substitute $x=7$ into the original equation: $5(7)=4(7)+7$. The left side is $35$ and the right side is $28+7=35$.
3. Check by substitution and state the result. Substituting $x=7$ gives $5(7)=35$ and $4(7)+7=35$, so the solution is $x=7$.

Takeaway: Subtracting the same variable term from both sides cancels that term without changing the equation's solution.

Example 8.23

Solution

1. Move the variable terms together. Subtract $5y$ from both sides of $5y-8=7y$: $5y-5y-8=7y-5y$, which gives $-8=2y$.
2. Apply the Division Property of Equality. Divide both sides by $2$: $\frac{-8}{2}=\frac{2y}{2}$, so $-4=y$.
3. Rewrite with the variable first. The equivalent form is $y=-4$.
4. Check by substitution. Substitute $y=-4$: $5(-4)-8=7(-4)$. The left side is $-20-8=-28$ and the right side is $-28$.
5. Check by substitution and state the result. Substituting $y=-4$ gives $5(-4)-8=-28$ and $7(-4)=-28$, so the solution is $y=-4$.

Takeaway: Subtracting equal variable terms from both sides preserves equality while making the equation much simpler.

Example 8.24

Solution

1. Eliminate the negative variable term. Add $x$ to both sides of $7x=-x+24$: $7x+x=-x+x+24$, which cancels $-x$ on the right without changing the solution.
2. Combine the like variable terms. On the left, $7x+x=8x$, while the right side becomes $24$, so the simplified equation is $8x=24$.
3. Apply the Division Property of Equality. Divide both sides by $8$: $\frac{8x}{8}=\frac{24}{8}$, and $24\div8=3$, so $x=3$.
4. Check by substitution and state the result. Substitute $x=3$ into the original equation: $7(3)=21$ and $-3+24=21$, so the solution is $x=3$.

Takeaway: Adding the opposite of a negative variable term cancels it and leaves a positive coefficient to divide.

Solve Equations with Variables and Constants on Both Sides

The next example will be the first to have variables and constants on both sides of the equation. As we did before, we’ll collect the variable terms to one side and the constants to the other side.

Example 8.25

Solution

1. Choose the variable side. The coefficients of the variable terms are $7$ and $6$, so keep $7x+5$ on the left as the variable side and move $6x$ from the right.
2. Collect the variable terms. Subtract $6x$ from both sides: $7x-6x+5=6x-6x+2$, and the variable terms on the right cancel.
3. Remove the constant. Subtract $5$ from both sides: $x+5-5=2-5$, which gives $x=-3$.
4. Check by substitution and state the result. Substitute $x=-3$: $7(-3)+5=-21+5=-16$ and $6(-3)+2=-18+2=-16$, so the solution is $x=-3$.

Takeaway: Subtracting $6x$ from both sides reduces the equation to $x+5=2$, so $x=-3$.

We’ll summarize the steps we took so you can easily refer to them.

It is a good idea to make the variable side the one in which the variable has the larger coefficient. This usually makes the arithmetic easier.

Example 8.26

Solution

1. Combine the variable terms. Add $3n$ to both sides of $6n-2=-3n+7$: $6n+3n-2=-3n+3n+7$, so $9n-2=7$.
2. Apply the Addition Property of Equality. Add $2$ to both sides: $9n-2+2=7+2$, which gives $9n=9$.
3. Apply the Division Property of Equality. Divide both sides by $9$: $\frac{9n}{9}=\frac{9}{9}$, so $n=1$.
4. Check by substitution and state the result. Substitute $n=1$: $6(1)-2=-3(1)+7$, so the solution is $n=1$.

Takeaway: Adding the magnitude of a negative variable coefficient to both sides combines the variables and keeps the equation balanced.

Example 8.27

Solution

1. Choose the variable side. The right side has the larger variable coefficient, so keep $5a+8$ on the right and subtract $2a$ from both sides of $2a-7=5a+8$.
2. Collect the variable terms. Subtract $2a$ from both sides: $2a-2a-7=5a-2a+8$, which gives $-7=3a+8$.
3. Remove the constant. Subtract $8$ from both sides: $-7-8=3a+8-8$, so $-15=3a$.
4. Apply the Division Property of Equality. Divide both sides by $3$: $\frac{-15}{3}=\frac{3a}{3}$, so $-5=a$.
5. Check by substitution and state the result. Substitute $a=-5$: $2(-5)-7=-17$ and $5(-5)+8=-17$, so the solution is $a=-5$.

Takeaway: Choosing the side with the larger variable coefficient keeps the algebra positive, and the equation gives $a=-5$.

To solve an equation with fractions, we still follow the same steps to get the solution.

Example 8.28

Solution

1. Combine the fractional variable terms. Subtract $\frac{1}{2}x$ from both sides of $\frac{3}{2}x+5=\frac{1}{2}x-3$: $x+5=-3$.
2. Apply the Subtraction Property of Equality. Subtract $5$ from both sides: $x+5-5=-3-5$, which gives $x=-8$.
3. Check by substitution and state the result. Substitute $x=-8$: $\frac{3}{2}(-8)+5=-12+5=-7$ and $\frac{1}{2}(-8)-3=-4-3=-7$, so the solution is $x=-8$.

Takeaway: Fractional coefficients can be combined like any other like terms before solving the remaining equation.

We follow the same steps when the equation has decimals, too.

Example 8.29

Solution

1. Combine the variable terms. Subtract $1.6x$ from both sides of $3.4x+4=1.6x-5$: $3.4x-1.6x+4=1.6x-1.6x-5$, giving $1.8x+4=-5$.
2. Apply the Subtraction Property of Equality. Subtract $4$ from both sides: $1.8x+4-4=-5-4$, so $1.8x=-9$.
3. Apply the Division Property of Equality. Divide both sides by $1.8$: $\frac{1.8x}{1.8}=\frac{-9}{1.8}$, which gives $x=-5$.
4. Check by substitution and state the result. Substitute $x=-5$: $3.4(-5)+4=-17+4=-13$ and $1.6(-5)-5=-8-5=-13$, so the solution is $x=-5$.

Takeaway: Treat decimal coefficients as exact numbers and check each division, since a misplaced decimal changes the value.

Solve Equations Using a General Strategy

Each of the first few sections of this chapter has dealt with solving one specific form of a linear equation. It’s time now to lay out an overall strategy that can be used to solve any linear equation. We call this the general strategy. Some equations won’t require all the steps to solve, but many will. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

Example 8.30

Solution

1. Distribute the factor. Expand $3(x+2)=18$ to obtain $3x+6=18$.
2. Apply the Subtraction Property of Equality. Subtract $6$ from both sides: $3x+6-6=18-6$, which gives $3x=12$.
3. Apply the Division Property of Equality. Divide both sides by $3$: $\frac{3x}{3}=\frac{12}{3}$, so $x=4$.
4. Check by substitution and state the result. Substitute $x=4$ into the original equation: $3(4+2)=3(6)=18$, so the solution is $x=4$.

Takeaway: Distribute before undoing constants, because a hidden product term can change the coefficient of the variable.

Example 8.31

Solution

1. Distribute the negative sign. Expand $-(x+5)=7$ as $-x-5=7$.
2. Apply the Addition Property of Equality. Add $5$ to both sides: $-x-5+5=7+5$, which gives $-x=12$.
3. Remove the negative coefficient. Divide both sides by $-1$: $\frac{-x}{-1}=\frac{12}{-1}$, so $x=-12$.
4. Check by substitution and state the result. Substitute $x=-12$ into the original equation: $-((-12)+5)=-(-7)=7$, so the solution is $x=-12$.

Takeaway: A negative sign outside parentheses changes the sign of every term inside, not just the first.

Example 8.32

Solution

1. Distribute the factor. Expand $4(x-2)+5=-3$ to obtain $4x-8+5=-3$.
2. Combine like constants. Since $-8+5=-3$, the equation becomes $4x-3=-3$.
3. Apply the Addition Property of Equality. Add $3$ to both sides: $4x-3+3=-3+3$, which gives $4x=0$.
4. Apply the Division Property of Equality. Divide both sides by $4$: $\frac{4x}{4}=\frac{0}{4}$, so $x=0$.
5. Check by substitution and state the result. Substitute $x=0$ into the original equation: $4(0-2)+5=-8+5=-3$, so the solution is $x=0$.

Takeaway: Combine constants only after distributing, because the constants inside and outside the parentheses combine with different signs.

Example 8.33

Solution

1. Distribute the negative factor. Expand $8-2(3y+5)=0$ as $8-6y-10=0$.
2. Combine like constants. The constants give $8-10=-2$, so the equation becomes $-6y-2=0$.
3. Apply the Addition Property of Equality. Add $2$ to both sides: $-6y-2+2=0+2$, which gives $-6y=2$.
4. Apply the Division Property of Equality. Divide both sides by $-6$: $\frac{-6y}{-6}=\frac{2}{-6}$, so $y=-\frac{1}{3}$.
5. Check by substitution and state the result. Substitute $y=-\frac{1}{3}$ into the original equation: $8-2(3(-\frac{1}{3})+5)=8-2(4)=0$, so the solution is $y=-\frac{1}{3}$.

Takeaway: Distribute a negative coefficient to every term inside the parentheses, then use the signs to combine constants correctly.

Example 8.34

Solution

1. Distribute both products. Expand $3(x-2)-5=4(2x+1)+5$ to obtain $3x-6-5=8x+4+5$.
2. Combine like terms on each side. The left side becomes $3x-11$, and the right side becomes $8x+9$, so $3x-11=8x+9$.
3. Collect the variable terms. Subtract $3x$ from both sides: $3x-3x-11=8x-3x+9$, which gives $-11=5x+9$.
4. Remove the constant. Subtract $9$ from both sides: $-11-9=5x+9-9$, so $-20=5x$.
5. Apply the Division Property of Equality. Divide both sides by $5$: $\frac{-20}{5}=\frac{5x}{5}$, giving $x=-4$.
6. Check by substitution. Substitute $x=-4$: $3(-4-2)-5=-18-5=-23$ and $4(2(-4)+1)+5=4(-7)+5=-23$. Since $-23=-23$, the solution is verified.

Takeaway: Simplify both sides fully before collecting variables, since an undistributed term can change the final coefficient.

Example 8.35

Solution

1. Distribute the fractional factor. Expand $\frac{1}{2}(6x-2)=5-x$ to obtain $3x-1=5-x$.
2. Collect the variable terms. Add $x$ to both sides: $3x+x-1=5-x+x$, which gives $4x-1=5$.
3. Remove the constant. Add $1$ to both sides: $4x-1+1=5+1$, so $4x=6$.
4. Apply the Division Property of Equality. Divide both sides by $4$: $\frac{4x}{4}=\frac{6}{4}$, so $x=\frac{3}{2}$.
5. Check by substitution and state the result. Substitute $x=\frac{3}{2}$ into the original equation: $\frac{1}{2}(9-2)=\frac{7}{2}$ and $5-\frac{3}{2}=\frac{7}{2}$, so the solution is $x=\frac{3}{2}$.

Takeaway: Distributing a fractional coefficient clears the fraction and makes variable collection straightforward.

In many applications, we will have to solve equations with decimals. The same general strategy will work for these equations.

Example 8.36

Solution

1. Distribute both decimal factors. Expand $0.24(100x+5)=0.4(30x+15)$ to obtain $24x+1.2=12x+6$.
2. Collect the variable terms. Subtract $12x$ from both sides: $24x+1.2-12x=12x+6-12x$, giving $12x+1.2=6$.
3. Remove the constant. Subtract $1.2$ from both sides: $12x+1.2-1.2=6-1.2$, so $12x=4.8$.
4. Apply the Division Property of Equality. Divide both sides by $12$: $\frac{12x}{12}=\frac{4.8}{12}$, which gives $x=0.4$.
5. Check by substitution and state the result. Substitute $x=0.4$ into the original equation: $0.24(40+5)=0.24(45)=10.8$ and $0.4(12+15)=0.4(27)=10.8$, so the solution is $x=0.4$.

Takeaway: Distributing decimal factors first turns the equation into one with integer coefficients and often simpler arithmetic.

Section 8.3 Exercises

Section 8.3 Exercises

Solve an Equation with Constants on Both Sides

In the following exercises, solve the equation for the variable.

Solve an Equation with Variables on Both Sides

In the following exercises, solve the equation for the variable.

Solve an Equation with Variables and Constants on Both Sides

In the following exercises, solve the equations for the variable.

Solve an Equation Using the General Strategy

In the following exercises, solve the linear equation using the general strategy.

Everyday Math

Writing Exercises

Self Check

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

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

1. Solve: $3x+4=-8.$

2. Solve: $5a+3=-37.$

3. Solve: $5y-9=16.$

4. Solve: $3m-8=19.$

5. Solve: $6n=5n+10.$

6. Solve: $-6c=-7c+1.$

1. Solve: $3p-14=5p.$

2. Solve: $8m+9=5m.$

3. Solve: $12j=-4j+32.$

4. Solve: $8h=-4h+12.$

5. Solve: $12x+8=6x+2.$

6. Solve: $9y+4=7y+12.$

1. Solve: $8q-5=-4q+7.$

2. Solve: $7n-3=n+3.$

3. Solve: $2a-2=6a+18.$

4. Solve: $4k-1=7k+17.$

5. Solve: $\frac{7}{8}\,x-12=-\frac{1}{8}\,x-2.$

6. Solve: $\frac{7}{6}\,y+11=\frac{1}{6}\,y+8.$

1. Solve: $2.8x+12=-1.4x-9.$

2. Solve: $3.6y+8=1.2y-4.$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

5. $7x-8=34$

6. $14y+7=91$

1. $4m+9=-23$

2. $-47=6b+1$

3. $29=-8x-3$

4. $-14q-15=13$

5. $9k=8k-11$

6. $6x+27=9x$

1. $b=-4b-15$

2. $7z=39-6z$

3. $8x+\frac{3}{4}=7x$

4. $-15r-8=-11r$

5. $4x-17=3x+2$

6. $21+6f=7f+14$

1. $8q-5=5q-20$

2. $9c+7=-2c-37$

3. $12x-17=-3x+13$

4. $3y-4=12-y$

5. $\frac{4}{3}\,m-7=\frac{1}{3}\,m-13$

6. $11-\frac{1}{4}\,a=\frac{3}{4}\,a+4$

1. $\frac{5}{4}\,a+15=\frac{3}{4}\,a-5$

2. $\frac{3}{5}\,p+2=\frac{4}{5}\,p-1$

3. $13z+6.45=8z+23.75$

4. $2.7w-80=1.2w+10$

5. $6.6x-18.9=3.4x+54.7$

6. $4(y+7)=64$

1. $9=3(x-3)$

2. $14(y-6)=-42$

3. $-7(3n+4)=14$

4. $8(3+3p)=0$

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

6. $4(2.5v-0.6)=7.6$

1. $0.5(16m+34)=-15$

2. $-(t-8)=17$

3. $8(6b-7)+23=63$

4. $13+2(m-4)=17$

5. $-9+6(5-k)=12$

6. $18-(9r+7)=-16$

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

2. $3(4n-1)-2=8n+3$

3. $5(x-4)-4x=14$

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

5. $4(x-1)-8=6(3x-2)-7$

6. Making a fence Jovani has a fence around the rectangular garden in his backyard. The perimeter of the fence is $150$ feet. The length is $15$ feet more than the width. Find the width, $w,$ by solving the equation $150=2(w+15)+2w.$

1. Coins Rhonda has $\text{\textdollar 1.90}$ 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 $0.05n+0.10(2n-1)=1.90.$

2. When solving an equation with variables on both sides, why is it usually better to choose the side with the larger coefficient as the variable side?

3. What is the first step you take when solving the equation $3-7(y-4)=38?$ Explain why this is your first step.

4. Using your own words, list the steps in the General Strategy for Solving Linear Equations.