8.4: Solve Equations with Variables and Constants on Both Sides (Part 1)
 Page ID
 5021
Skills to Develop
 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
be prepared!
Before you get started, take this readiness quiz.
 Simplify: 4y − 9 + 9. If you missed this problem, review Example 2.3.10.
 Solve: y + 12 = 16. If you missed this problem, review Example 2.5.4.
 Solve: −3y = 63. If you missed this problem, review Example 3.9.6.
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 \(\PageIndex{1}\):
Solve: 4x + 6 = −14.
Solution
In this equation, the variable is only on the left side. It makes sense to call the left side the variable side. Therefore, the right side will be the constant side. We’ll write the labels above the equation to help us remember what goes where.
Since the left side is the variable side, the 6 is out of place. We must "undo" adding 6 by subtracting 6, and to keep the equality we must subtract 6 from both sides. Use the Subtraction Property of Equality.  $$4x + 6 \textcolor{red}{6} = 14 \textcolor{red}{6}$$ 
Simplify.  $$4x = 20$$ 
Now all the x's are on the left and the constant on the right.  
Use the Division Property of Equality.  $$\dfrac{4x}{\textcolor{red}{4}} = \dfrac{20}{\textcolor{red}{4}}$$ 
Simplify.  $$x = 5$$ 
Check: Let x = −5.  $$\begin{split} 4x + 6 &= 14 \\ 4(\textcolor{red}{5}) + 6 &= 14 \\ 20 + 6 &= 14 \\ 14 &= 14\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{1}\):
Solve: 3x + 4 = −8.
 Answer

x = 4
Exercise \(\PageIndex{2}\):
Solve: 5a + 3 = −37.
 Answer

a = 8
Example \(\PageIndex{2}\):
Solve: 2y − 7 = 15.
Solution
Notice that the variable is only on the left side of the equation, so this will be the variable side and the right side will be the constant side. Since the left side is the variable side, the 7 is out of place. It is subtracted from the 2y, so to ‘undo’ subtraction, add 7 to both sides.
Add 7 to both sides.  $$2y  7 \textcolor{red}{+7} = 15 \textcolor{red}{+7}$$ 
Simplify.  $$2y = 22$$ 
The variables are now on one side and the constants on the other.  
Divide both sides by 2.  $$\dfrac{2y}{\textcolor{red}{2}} = \dfrac{22}{\textcolor{red}{2}}$$ 
Simplify.  $$y = 11$$ 
Check: Substitute: y = 11.  $$\begin{split} 2y  7 &= 15 \\ 2 \cdot \textcolor{red}{11}  7 &\stackrel{?}{=} 15 \\ 22  7 &\stackrel{?}{=} 15 \\ 15 &= 15\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{3}\):
Solve: 5y − 9 = 16.
 Answer

y = 5
Exercise \(\PageIndex{4}\):
Solve: 3m − 8 = 19.
 Answer

m = 9
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 \(\PageIndex{3}\):
Solve: 5x = 4x + 7.
Solution
Here the variable, x, is on both sides, but the constants appear only on the right side, so let’s make the right side the “constant” side. Then the left side will be the “variable” side.
We don't want any variables on the right, so subtract the 4x.  $$5x \textcolor{red}{4x} = 4x \textcolor{red}{4x} + 7$$ 
Simplify.  $$x = 7$$ 
We have all the variables on one side and the constants on the other. We have solved the equation.  
Check: Substitute 7 for x.  $$\begin{split} 5x &= 4x + 7 \\ 5(\textcolor{red}{7}) &\stackrel{?}{=} 4(\textcolor{red}{7}) + 7 \\ 35 &\stackrel{?}{=} 28 + 7 \\ 35 &= 35\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{5}\):
Solve: 6n = 5n + 10.
 Answer

n = 10
Exercise \(\PageIndex{6}\):
Solve: −6c = −7c + 1.
 Answer

c = 1
Example \(\PageIndex{4}\):
Solve: 5y − 8 = 7y.
Solution
The only constant, −8, is on the left side of the equation and variable, y, is on both sides. Let’s leave the constant on the left and collect the variables to the right.
Subtract 5y from both sides.  $$5y \textcolor{red}{5y} 8 = 7y \textcolor{red}{5y}$$ 
Simplify.  $$8 = 2y$$ 
We have the variables on the right and the constants on the left. Divide both sides by 2.  $$\dfrac{8}{\textcolor{red}{2}} = \dfrac{2y}{\textcolor{red}{2}}$$ 
Simplify.  $$4 = y$$ 
Rewrite with the variable on the left.  $$y = 4$$ 
Check: Let y = −4.  $$\begin{split} 5y  8 &= 7y \\ 5(\textcolor{red}{4}) 8 &\stackrel{?}{=} 7(\textcolor{red}{4}) \\ 20  8 &\stackrel{?}{=} 28 \\ 28 &= 28\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{7}\):
Solve: 3p − 14 = 5p.
 Answer

p = 7
Exercise \(\PageIndex{8}\):
Solve: 8m + 9 = 5m.
 Answer

m = 3
Example \(\PageIndex{5}\):
Solve: 7x = − x + 24.
Solution
The only constant, 24, is on the right, so let the left side be the variable side.
Remove the −x from the right side by adding x to both sides.  $$7x \textcolor{red}{+x} = x \textcolor{red}{+x} + 24$$ 
Simplify.  $$8x = 24$$ 
All the variables are on the left and the constants are on the right. Divide both sides by 8.  $$\dfrac{8x}{\textcolor{red}{8}} = \dfrac{24}{\textcolor{red}{8}}$$ 
Simplify.  $$x = 3$$ 
Check: Substitute x = 3.  $$\begin{split} 7x &= x + 24 \\ 7(\textcolor{red}{3}) &\stackrel{?}{=} (\textcolor{red}{3}) + 24 \\ 21 &= 21\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{9}\):
Solve: 12j = −4j + 32.
 Answer

j = 2
Exercise \(\PageIndex{10}\):
Solve: 8h = −4h + 12.
 Answer

h = 1
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 \(\PageIndex{6}\):
Solve: 7x + 5 = 6x + 2.
Solution
Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are 7x and 6x. Since 7 is greater than 6, make the left side the variable side and so the right side will be the constant side.
Collect the variable terms to the left side by subtracting 6x from both sides.  $$7x \textcolor{red}{6x} + 5 = 6x \textcolor{red}{6x} +2$$ 
Simplify.  $$x + 5 = 2$$ 
Now, collect the constants to the right side by subtracting 5 from both sides.  $$x + 5 \textcolor{red}{5} = 2 \textcolor{red}{5}$$ 
Simplify.  $$x = 3$$ 
The solution is x = −3.  
Check: Let x = −3.  $$\begin{split} 7x + 5 &= 6x + 2 \\ 7(\textcolor{red}{3}) + 5 &\stackrel{?}{=} 6(\textcolor{red}{3}) + 2 \\ 21 + 5 &\stackrel{?}{=} 18 + 2 \\ 16 &= 16\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{11}\):
Solve: 12x + 8 = 6x + 2.
 Answer

x = 1
Exercise \(\PageIndex{12}\):
Solve: 9y + 4 = 7y + 12.
 Answer

y = 4
We’ll summarize the steps we took so you can easily refer to them.
HOW TO: SOLVE AN EQUATION WITH VARIABLES AND CONSTANTS ON BOTH SIDES
Step 1. Choose one side to be the variable side and then the other will be the constant side.
Step 2. Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.
Step 3. Collect the constants to the other side, using the Addition or Subtraction Property of Equality.
Step 4. Make the coefficient of the variable 1, using the Multiplication or Division Property of Equality.
Step 5. Check the solution by substituting it into the original equation.
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 \(\PageIndex{7}\):
Solve: 6n − 2 = −3n + 7.
Solution
We have 6n on the left and −3n on the right. Since 6 > − 3, make the left side the “variable” side.
We don't want variables on the right side—add 3n to both sides to leave only constants on the right.  $$6n \textcolor{red}{+3n}  2 = 3n \textcolor{red}{+3n} +7$$ 
Combine like terms.  $$9n  2 = 7$$ 
We don't want any constants on the left side, so add 2 to both sides.  $$9n  2 \textcolor{red}{+2} = 7 \textcolor{red}{+2}$$ 
Simplify.  $$9n = 9$$ 
The variable term is on the left and the constant term is on the right. To get the coefficient of n to be one, divide both sides by 9.  $$\dfrac{9n}{\textcolor{red}{9}} = \dfrac{9}{\textcolor{red}{9}}$$ 
Simplify.  $$n = 1$$ 
Check: Substitute 1 for n.  $$\begin{split} 6n  2 &= 3n + 7 \\ 6(\textcolor{red}{1})  2 &\stackrel{?}{=} + 7 \\ 4 &= 4\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{13}\):
Solve: 8q − 5 = −4q + 7.
 Answer

q = 1
Exercise \(\PageIndex{14}\):
Solve: 7n − 3 = n + 3.
 Answer

n = 1
Example \(\PageIndex{8}\):
Solve: 2a − 7 = 5a + 8.
Solution
This equation has 2a on the left and 5a on the right. Since 5 > 2, make the right side the variable side and the left side the constant side.
Subtract 2a from both sides to remove the variable term from the left.  $$2a \textcolor{red}{2a}  7 = 5a \textcolor{red}{2a} + 8$$ 
Combine like terms.  $$7 = 3a + 8$$ 
Subtract 8 from both sides to remove the constant from the right.  $$7 \textcolor{red}{8} = 3a + 8 \textcolor{red}{8}$$ 
Simplify.  $$15 = 3a$$ 
Divide both sides by 3 to make 1 the coefficient of a.  $$\dfrac{15}{\textcolor{red}{3}} = \dfrac{3a}{\textcolor{red}{3}}$$ 
Simplify.  $$5 = a$$ 
Check: Let a = −5.  $$\begin{split} 2a  7 &= 5a + 8 \\ 2(\textcolor{red}{5})  7 &\stackrel{?}{=} 5(\textcolor{red}{5}) + 8 \\ 10  7 &\stackrel{?}{=} 25 + 8 \\ 17 &= 17\; \checkmark \end{split}$$ 
Note that we could have made the left side the variable side instead of the right side, but it would have led to a negative coefficient on the variable term. While we could work with the negative, there is less chance of error when working with positives. The strategy outlined above helps avoid the negatives!
Exercise \(\PageIndex{15}\):
Solve: 2a − 2 = 6a + 18.
 Answer

a = 5
Exercise \(\PageIndex{16}\):
Solve: 4k − 1 = 7k + 17.
 Answer

k = 6
To solve an equation with fractions, we still follow the same steps to get the solution.
Example \(\PageIndex{9}\):
Solve: \(\dfrac{3}{2}\)x + 5 = \(\dfrac{1}{2}\)x − 3.
Solution
Since \(\dfrac{3}{2} > \dfrac{1}{2}\), make the left side the variable side and the right side the constant side.
Subtract \(\dfrac{1}{2}\)x from both sides.  $$\dfrac{3}{2} x \textcolor{red}{ \dfrac{1}{2} x} + 5 = \dfrac{1}{2} x \textcolor{red}{\dfrac{1}{2} x}  3$$ 
Combine like terms.  $$x + 5 = 3$$ 
Subtract 5 from both sides.  $$x + 5 \textcolor{red}{5} = 3 \textcolor{red}{5}$$ 
Simplify.  $$x = 8$$ 
Check: Let x = −8.  $$\begin{split} \dfrac{3}{2} x + 5 &= \dfrac{1}{2} x  3 \\ \dfrac{3}{2} (\textcolor{red}{8}) + 5 &\stackrel{?}{=} \dfrac{1}{2} (\textcolor{red}{8})  3 \\ 12 + 5 &\stackrel{?}{=} 4  3 \\ 7 &= 7\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{17}\):
Solve: \(\dfrac{7}{8}\)x  12 = \( \dfrac{1}{8}\)x − 2.
 Answer

x = 10
Exercise \(\PageIndex{18}\):
Solve: \(\dfrac{7}{6}\)y + 11 = \(\dfrac{1}{6}\)y + 8.
 Answer

y = 3
We follow the same steps when the equation has decimals, too.
Example \(\PageIndex{10}\):
Solve: 3.4x + 4 = 1.6x − 5.
Solution
Since 3.4 > 1.6, make the left side the variable side and the right side the constant side.
Subtract 1.6x from both sides.  $$3.4x \textcolor{red}{1.6x} + 4 = 1.6x \textcolor{red}{1.6x}  5$$ 
Combine like terms.  $$1.8x + 4 = 5$$ 
Subtract 4 from both sides.  $$1.8x + 4 \textcolor{red}{4} = 5 \textcolor{red}{4}$$ 
Simplify.  $$1.8x = 9$$ 
Use the Division Property of Equality.  $$\dfrac{1.8x}{\textcolor{red}{1.8}} = \dfrac{9}{\textcolor{red}{1.8}}$$ 
Simplify.  $$x = 5$$ 
Check: Let x = −5.  $$\begin{split} 3.4x + 4 &= 1.6x  5 \\ 3.4(\textcolor{red}{5}) + 4 &\stackrel{?}{=} 1.6(\textcolor{red}{5})  5 \\ 17 + 4 &\stackrel{?}{=} 8  5 \\ 13 &= 13\; \checkmark \end{split}$$ 
Exercise \(\PageIndex{19}\):
Solve: 2.8x + 12 = −1.4x − 9.
 Answer

x = 5
Exercise \(\PageIndex{20}\):
Solve: 3.6y + 8 = 1.2y − 4.
 Answer

y = 5
Contributors and Attributions
Lynn Marecek (Santa Ana College) and MaryAnne AnthonySmith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/fd53eae1fa2...49835c3c@5.191."