Solving Linear Equations that Require Simplification
Learning Outcomes
- Solve equations that need to be simplified
- Solve a linear equation that requires multiple steps and a combination of the properties of equality
- Use the distributive property to solve equations containing parentheses
Using the Properties of Equality to Solve Linear Equations
Some linear equations can be solved in one or two steps, while others are more complicated. Although multi-step equations take more time and more operations to solve, they can still be simplified and solved by applying basic algebraic rules. In this section, we will look at equations that require some additional steps before they can be solved.
Combining Like Terms
Many equations start out more complicated than the ones we’ve just solved in the previous section. Let’s work through some examples that will employ simplifying by combining like terms.
Example
Solve: [latex]8x+9x - 5x=-3+15[/latex]
Solution:
First, we need to simplify both sides of the equation as much as possible.
Start by combining like terms to simplify each side.
| [latex]8x+9x-5x=-3+15[/latex] | |
| Combine like terms. | [latex]12x=12[/latex] |
| Divide both sides by 12 to isolate x. | [latex]\Large\frac{12x}{\color{red}{12}}\normalsize =\Large\frac{12}{\color{red}{12}}[/latex] |
| Simplify. | [latex]x=1[/latex] |
| Check your answer. Let [latex]x=1[/latex] | |
| [latex]8x+9x-5x=-3+15[/latex] | |
| [latex]8\cdot\color{red}{1}+9\cdot\color{red}{1}-5\cdot\color{red}{1}\stackrel{\text{?}}{=}-3+15[/latex] | |
| [latex]8+9-5\stackrel{\text{?}}{=}-3+15[/latex] | |
| [latex]12=12\quad\checkmark[/latex] |
Example
Solve [latex]3x+5x+4-x+7=88[/latex]
Try it
The video shows an example of solving a linear equation that requires combining like terms.
Variables on the right side of the equation
It is not necessary to always have the variables on the left side of the equation. The Property of Reflection tells us that, if [latex]a=b[/latex], then [latex]b=a[/latex]. Let’s consider an example with variables on the right side.
example
Solve: [latex]11 - 20=17y - 8y - 6y[/latex]
Solution:
Simplify each side by combining like terms.
| [latex]11-20=17y-8y-6y[/latex] | |
| Simplify each side. | [latex]-9=3y[/latex] |
| Divide both sides by 3 to isolate y. | [latex]\Large\frac{-9}{\color{red}{3}}\normalsize =\Large\frac{3y}{\color{red}{3}}[/latex] |
| Simplify. | [latex]-3=y[/latex] |
| Check your answer. Let [latex]y=-3[/latex] | |
| [latex]11-20=17y-8y-6y[/latex] | |
| [latex]11-20\stackrel{\text{?}}{=}17( \color{red}{-3})-8(\color{red}{-3})-6(\color{red}{-3})[/latex] | |
| [latex]11-20\stackrel{\text{?}}{=}-51+24+18[/latex] | |
| [latex]-9=-9\quad\checkmark[/latex] |
Notice that the variable ended up on the right side of the equals sign when we solved the equation. It is usually preferable to take one more step to write the solution with the variable on the left side of the equals sign.
try it
Variables on both sides of the equation
So far we have solved equations with variables on only one side of the equation. This does not happen all the time—so now we’ll see how to solve equations where there are variable terms on both sides of the equation, as in this equation: [latex]4x-6=2x+10[/latex]. We will start by 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 we do to the left side of the equation, we must do to the right side as well.
To solve [latex]4x-6=2x+10[/latex], we need to “move” one of the variable terms. This can make it difficult to decide which side to work with. It doesn’t matter which term gets moved, [latex]4x[/latex] or [latex]2x[/latex], however, to avoid negative coefficients, we can move the smaller term.
Examples
Solve: [latex]4x-6=2x+10[/latex]
Solution
Choose the variable term to “move”—to avoid negative terms subtract [latex]2x[/latex] from both sides:
[latex]\,\,\,4x-6=2x+10\\\underline{\color{blue}{-2x}\,\,\,\,\,\,\,\,\,\,\color{blue}{-2x}}\\\,\,\,2x-6=10[/latex]
Add 6 to both sides to isolate the variable term:
[latex]\begin{array}{r}2x-6=10\\\underline{\,\,\,\,\color{blue}{+6}\,\,\,\color{blue}{+6}}\\2x=16\end{array}[/latex]
Divide each side by [latex]2[/latex] to isolate the variable [latex]x[/latex].
[latex]\begin{array}{c}\frac{2x}{\color{blue}{2}}=\frac{16}{\color{blue}{2}}\\\\x=8\end{array}[/latex]
Check the solution: [latex]4x-6=2x+10 \\ 4\color{blue}{(8)}-6 = 2\color{blue}{(8)}+10 \\ 32-6=16+10 \\ 26=26[/latex] True
Answer
[latex]x=8[/latex]
This video shows an example of solving equations that have variables on both sides of the equals sign.
In the next example, the variable, [latex]x[/latex], is on both sides, but the constants appear only on the right side, so we’ll make the right side the “constant” side. Then the left side will be the “variable” side.
ExampleS
Solve:
1. [latex]5x=4x+7[/latex]
2. [latex]7x=-x+24[/latex]
Solution 1:
| [latex]5x[/latex] is the side containing only a variable.[latex]4x+7[/latex] is the side containing a constant. | ||
| We don’t want any variables on the right, so subtract the [latex]4x[/latex] . | [latex]5x\color{red}{-4x}=4x\color{red}{-4x}+7[/latex] | |
| Simplify. | [latex]x=7[/latex] | |
| We have all the variables on one side and the constants on the other. We have solved the equation. | ||
| Check: | [latex]5x=4x+7[/latex] | |
| Substitute [latex]7[/latex] for [latex]x[/latex] . | [latex]5(\color{red}{7})\stackrel{\text{?}}{=}4(\color{red}{7})+7[/latex] | |
| [latex]35\stackrel{\text{?}}{=}28+7[/latex] | ||
| [latex]35=35\quad\checkmark[/latex] | ||
Solution 2:
The only constant, [latex]24[/latex], is on the right, so let the left side be the variable side.
| [latex]7x[/latex] is the side containing only a variable.[latex]-x+24[/latex] is the side containing a constant. | |
| Remove the [latex]-x[/latex] from the right side by adding [latex]x[/latex] to both sides. | [latex]7x\color{red}{+x}=-x\color{red}{+x}+24[/latex] |
| Simplify. | [latex]8x=24[/latex] |
| All the variables are on the left and the constants are on the right. Divide both sides by [latex]8[/latex]. | [latex]\Large\frac{8x}{\color{red}{8}}\normalsize =\Large\frac{24}{\color{red}{8}}[/latex] |
| Simplify. | [latex]x=3[/latex] |
| Check: | [latex]7x=-x+24[/latex] |
| Substitute [latex]x=3[/latex]. | [latex]7(\color{red}{3})\stackrel{\text{?}}{=}-(\color{red}{3})+24[/latex] |
| [latex]21=21\quad\checkmark[/latex] |
try it
Sometimes it is beneficial to have the variable term on the right side of the equation. There is no “correct” side to have the variable term, but the choice can help to avoid working with negative signs.
example
Solve: [latex]5y - 8=7y[/latex]
Solution:
The only constant, [latex]-8[/latex], is on the left side of the equation, and the variable, [latex]y[/latex], is on both sides. Let’s leave the constant on the left and collect the variables to the right.
| [latex]5y-8[/latex] is the side containing a constant.[latex]7y[/latex] is the side containing only a variable. | |
| Subtract [latex]5y[/latex] from both sides. | [latex]5y\color{red}{-5y}-8=7y\color{red}{-5y}[/latex] |
| Simplify. | [latex]-8=2y[/latex] |
| We have the variables on the right and the constants on the left. Divide both sides by [latex]2[/latex]. | [latex]\Large\frac{-8}{\color{red}{2}}\normalsize =\Large\frac{2y}{\color{red}{2}}[/latex] |
| Simplify. | [latex]-4=y[/latex] |
| Rewrite with the variable on the left. | [latex]y=-4[/latex] |
| Check: | [latex]5y-8=7y[/latex] |
| Let [latex]y=-4[/latex]. | [latex]5(\color{red}{-4})-8\stackrel{\text{?}}{=}7(\color{red}{-4})[/latex] |
| [latex]-20-8\stackrel{\text{?}}{=}-28[/latex] | |
| [latex]-28=-28\quad\checkmark[/latex] |
[/hidden-answer]
try it
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. As the number of variable and constant terms increase, so do the number of steps it takes to solve the equation.
Examples
Solve:
1. [latex]7x+5=6x+2[/latex]
2. [latex]6n - 2=-3n+7[/latex]
1. Solution:
Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are [latex]7x[/latex] and [latex]6x[/latex]. Since [latex]7[/latex] is greater than [latex]6[/latex], make the left side the variable side, and so the right side will be the constant side.
| [latex]7x+5=6x+2[/latex] | |
| Collect the variable terms to the left side by subtracting [latex]6x[/latex] from both sides. | [latex]7x\color{red}{-6x}+5=6x\color{red}{-6x}+2[/latex] |
| Simplify. | [latex]x+5=2[/latex] |
| Now, collect the constants to the right side by subtracting [latex]5[/latex] from both sides. | [latex]x+5\color{red}{-5}=2\color{red}{-5}[/latex] |
| Simplify. | [latex]x=-3[/latex] |
| The solution is [latex]x=-3[/latex] . | |
| Check: | [latex]7x+5=6x+2[/latex] |
| Let [latex]x=-3[/latex]. | [latex]7(\color{red}{-3})+5\stackrel{\text{?}}{=}6(\color{red}{-3})+2[/latex] |
| [latex]-21+5\stackrel{\text{?}}{=}-18+2[/latex] | |
| [latex]16=16\quad\checkmark[/latex] |
Answer: [latex]x=3[/latex]
2. Solution
Solve: [latex]6n - 2=-3n+7[/latex]
We have [latex]6n[/latex] on the left and [latex]-3n[/latex] on the right. Since [latex]6>-3[/latex], make the left side the “variable” side.
| [latex]6n-2=-3n+7[/latex] | |
| We don’t want variables on the right side—add [latex]3n[/latex] to both sides to leave only constants on the right. | [latex]6n\color{red}{+3n}-2=-3n\color{red}{+3n}+7[/latex] |
| Combine like terms. | [latex]9n-2=7[/latex] |
| We don’t want any constants on the left side, so add [latex]2[/latex] to both sides. | [latex]9n-2\color{red}{+2}=7\color{red}{+2}[/latex] |
| Simplify. | [latex]9n=9[/latex] |
| The variable term is on the left and the constant term is on the right. To get the coefficient of [latex]n[/latex] to be one, divide both sides by [latex]9[/latex]. | [latex]\Large\frac{9n}{\color{red}{9}}\normalsize =\Large\frac{9}{\color{red}{9}}[/latex] |
| Simplify. | [latex]n=1[/latex] |
| Check: | [latex]6n-2=-3n+7[/latex] |
| Substitute [latex]1[/latex] for [latex]n[/latex]. | [latex]6(\color{red}{1})-2\stackrel{\text{?}}{=}-3(\color{red}{1})+7[/latex] |
| [latex]4=4\quad\checkmark[/latex] |
Answer: [latex]n=1[/latex]
The following video shows an example of how to solve a multi-step equation. It doesn’t matter which side we choose to be the variable side; we can get the correct answer either way.
In the next example, we move the variable terms to the right side, to keep a positive coefficient on the variable.
EXAMPLE
Solve: [latex]2a - 7=5a+8[/latex]
Solution:
This equation has [latex]2a[/latex] on the left and [latex]5a[/latex] on the right. Since [latex]5>2[/latex], make the right side the variable side and the left side the constant side.
| [latex]2a-7=5a+8[/latex] | |
| Subtract [latex]2a[/latex] from both sides to remove the variable term from the left. | [latex]2a\color{red}{-2a}-7=5a\color{red}{-2a}+8[/latex] |
| Combine like terms. | [latex]-7=3a+8[/latex] |
| Subtract [latex]8[/latex] from both sides to remove the constant from the right. | [latex]-7\color{red}{-8}=3a+8\color{red}{-8}[/latex] |
| Simplify. | [latex]-15=3a[/latex] |
| Divide both sides by [latex]3[/latex] to make [latex]1[/latex] the coefficient of [latex]a[/latex] . | [latex]\Large\frac{-15}{\color{red}{3}}\normalsize =\Large\frac{3a}{\color{red}{3}}[/latex] |
| Simplify. | [latex]-5=a[/latex] |
| Check: | [latex]2a-7=5a+8[/latex] |
| Let [latex]a=-5[/latex] | [latex]2(\color{red}{-5})-7\stackrel{\text{?}}{=}5(\color{red}{-5})+8[/latex] |
| [latex]-10-7\stackrel{\text{?}}{=}-25+8[/latex] | |
| [latex]-17=-17\quad\checkmark[/latex] |
The following video shows another example of solving a multi-step equation.
try it
We just showed a lot of examples of different kinds of linear equations that may be encountered. There are some good habits to develop that will help us solve all kinds of linear equations. We’ll summarize the steps we took for easy reference.
Solve an equation with variables and constants on both sides
- Choose one side to be the variable side, and the other to be the constant side.
- Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.
- Collect the constants to the other side, using the Addition or Subtraction Property of Equality.
- Make the coefficient of the variable [latex]1[/latex], using the Multiplication or Division Property of Equality.
- Check the solution by substituting it into the original equation.
No Solution and Infinite Solutions
Equations that have a solution are called conditional equations. The truth of the equation is found only when the solution is entered in the equation. For example, [latex]2x-5=3[/latex] is true only when the solution [latex]x=4[/latex] is put into the equation. Any other value of [latex]x[/latex] makes the equation false. The truth of the equation is conditional upon the value of the variable.
It is possible for a linear equation in one variable to have an infinite number of solutions because it is always true. Equations that are always true are called identities. For example the equation [latex]3x+5x=8x[/latex] is always true, no matter the value of [latex]x[/latex]. Consequently, the equation has an infinite number of solutions. All real numbers are solutions of this equation.
It is also possible for a linear equation to have no solution. Such equations are called contradictions. For example, [latex]3x+5=3x-2[/latex] simplifies to [latex]5=-2[/latex], which is a contradiction. Consequently, this equation has no solution.
EXAMPLE
Solve the equation:
1. [latex]4x-3=2x+5+2x[/latex]
2. [latex]5x-4-3x=2x-4[/latex]
Solution:
| 1. | [latex]4x-3=2x+5+2x[/latex] |
| Simplify the right side by adding like terms | [latex]4x-3=4x+5[/latex] |
| Subtract [latex]4x[/latex] from both sides | [latex]-3=5[/latex] |
| Contradiction | No solution |
| 2. | [latex]5x-4-3x=2x-4[/latex] |
| Simplify the right side by adding like terms | [latex]2x-4=2x-4[/latex] |
| Subtract [latex]4x[/latex] from both sides | [latex]-4=-4[/latex] |
| Identity | Solution is all real numbers |
TRY IT
Solve the equation.
1. [latex]5x+2=5x[/latex]
2. [latex]3x-4+9=5x+3-2x+2[/latex]
When solving a linear equation in one variable, if we end up with a statement that is always true, we have an identity. A identity is true no matter the value of the variable, so the solution is the set of all real numbers. On the other hand, if we end up with an statement that is always false, we have a contradiction. A contradiction is never true, so the equation has no solution.
The Distributive Property
As we solve linear equations, we often need to do some work to write the linear equations in a form we are familiar with solving. This section will focus on manipulating an equation we are asked to solve in such a way that we can use the skills we learned for solving multi-step equations to ultimately arrive at the solution.
The distributive property can come in handy to simplify an equation. Using this property, we multiply the number in front of the parentheses by each term inside of the parentheses.
The Distributive Property of Multiplication over addition
For all real numbers [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex], [latex]a(b+c)=ab+ac[/latex].
When a number multiplies an expression inside parentheses, we can distribute the multiplication to each term of the expression individually.
Example
Solve for [latex]a[/latex].
[latex]4\left(2a+3\right)=28[/latex]
Solution:
Apply the distributive property to expand [latex]4\left(2a+3\right)[/latex] to [latex]8a+12[/latex]
[latex]\begin{array}{r}4\left(2a+3\right)=28\\ 8a+12=28\end{array}[/latex]
Subtract [latex]12[/latex] from both sides to isolate the variable term.
[latex]\begin{array}{r}8a+12\,\,\,=\,\,\,28\\ \underline{-12\,\,\,\,\,\,-12}\\ 8a\,\,\,=\,\,\,16\end{array}[/latex]
Divide both terms by [latex]8[/latex] to get a coefficient of [latex]1[/latex].
[latex]\begin{array}{r}\underline{8a}=\underline{16}\\8\,\,\,\,\,\,\,\,\,\,\,\,8\\a\,=\,\,2\end{array}[/latex]
Answer
[latex]a=2[/latex]
In our next example, we will use the distributive property of multiplication over addition first, simplify, then use the division property to finally solve.
example
Solve: [latex]-3\left(n - 2\right)-6=21[/latex]
Solution:
| [latex]-3(n-2)-6=21[/latex] | |
| Distribute. | [latex]-3n+6-6=21[/latex] |
| Simplify. | [latex]-3n=21[/latex] |
| Divide both sides by -3 to isolate n. | [latex]\Large\frac{-3n}{\color{red}{-3}}\normalsize =\Large\frac{21}{\color{red}{-3}}[/latex][latex]n=-7[/latex] |
| Check your answer. Let [latex]n=-7[/latex] . | |
| [latex]-3(n-2)-6=21[/latex] | |
| [latex]-3(\color{red}{-7}-2)-6\stackrel{\text{?}}{=}21[/latex] | |
| [latex]-3(-9)-6\stackrel{\text{?}}{=}21[/latex] | |
| [latex]27-6\stackrel{\text{?}}{=}21[/latex] | |
| [latex]21=21\quad\checkmark[/latex] |
try it
The video that follows shows another example of how to use the distributive property to solve a multi-step linear equation.
example
Solve: [latex]3\left(n - 4\right)-2n=-3[/latex]
Solution:
The left side of the equation has an expression that we should simplify.
| [latex]3(n-4)-2n=-3[/latex] | |
| Distribute on the left. | [latex]3n-12-2n=-3[/latex] |
| Use the Commutative Property to rearrange terms. | [latex]3n-2n-12=-3[/latex] |
| Combine like terms. | [latex]n-12=-3[/latex] |
| Isolate n using the Addition Property of Equality. | [latex]n-12\color{red}{+12}=-3\color{red}{+12}[/latex] |
| Simplify. | [latex]n=9[/latex] |
| Check.Substitute [latex]n=9[/latex] into the original equation. [latex]3(n-4)-2n=-3[/latex] [latex]3(\color{red}{9}-4)-2\cdot\color{red}{9}=-3[/latex] [latex]3(5)-18=-3[/latex] [latex]15-18=-3[/latex] [latex]-3=-3\quad\checkmark[/latex] The solution checks. |
TRY IT
The next example has expressions on both sides that need to be simplified.
example
Solve: [latex]2\left(3k - 1\right)-5k=-2 - 7[/latex]
Solution:
Both sides of the equation have expressions that we should simplify before we isolate the variable.
| [latex]2(3k-1)-5k=-2-7[/latex] | |
| Distribute on the left, subtract on the right. | [latex]6k-2-5k=-9[/latex] |
| Use the Commutative Property of Addition. | [latex]6k-5k-2=-9[/latex] |
| Combine like terms. | [latex]k-2=-9[/latex] |
| Undo subtraction by using the Addition Property of Equality. | [latex]k-2\color{red}{+2}=-9\color{red}{+2}[/latex] |
| Simplify. | [latex]k=-7[/latex] |
| Check.Let [latex]k=-7[/latex]. [latex]2(3k-1)-5k=-2-7[/latex] [latex]2(3(\color{red}{-7}-1)-5(\color{red}{-7})=-2-7[/latex] [latex]2(-21-1)-5(-7)=-9[/latex] [latex]2(-22)+35=-9[/latex] [latex]-44+35=-9[/latex] [latex]-9=-9\quad\checkmark[/latex] |
The solution checks.
Answer: [latex]k=-7[/latex]
TRY IT
The following video presents another example of how to solve an equation that requires simplifying before using the addition and subtraction properties.
In the next example, there are parentheses on both sides of the equals sign, so we need to use the distributive property twice.
Example
Solve for [latex]t[/latex].
[latex]2\left(4t-5\right)=-3\left(2t+1\right)[/latex]
Solution:
Apply the distributive property to expand [latex]2\left(4t-5\right)[/latex] to [latex]8t-10[/latex] and [latex]-3\left(2t+1\right)[/latex] to [latex]-6t-3[/latex]. Be careful in this step—you are distributing a negative number, so keep track of the sign of each number after you multiply.
[latex]\begin{array}{r}2\left(4t-5\right)=-3\left(2t+1\right)\,\,\,\,\,\, \\ 8t-10=-6t-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Add [latex]-6t[/latex] to both sides to begin combining like terms.
[latex]\begin{array}{r}8t-10=-6t-3\\ \underline{+6t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6t}\,\,\,\,\,\,\,\\ 14t-10=\,\,\,\,-3\,\,\,\,\,\,\,\end{array}[/latex]
Add [latex]10[/latex] to both sides of the equation to isolate [latex]t[/latex].
[latex]\begin{array}{r}14t-10=-3\\ \underline{+10\,\,\,+10}\\ 14t=\,\,\,7\,\end{array}[/latex]
The last step is to divide both sides by [latex]14[/latex] to completely isolate [latex]t[/latex].
[latex]\begin{array}{r}14t=7\,\,\,\,\\\frac{14t}{14}=\frac{7}{14}\end{array}[/latex]
Answer
[latex]t=\frac{1}{2}[/latex]
We simplified the fraction [latex]\frac{7}{14}[/latex] into [latex]\frac{1}{2}[/latex]
Try It
The following video solves another multi-step equation with two sets of parentheses.
