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# Solve Equations with the Distributive Property

### Learning Outcomes

• Use the properties of equality and the distributive property to solve equations containing parentheses
• Clear fractions and decimals from equations to make them easier to solve

## 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. If an equation you encounter contains parentheses, you'll need to clear them from the equations before attempting to combine like terms. To clear parentheses from an equation, use the distributive property to multiply the number in front of the parentheses by each term inside of the parentheses. If the number in front of the parentheses is negative, multiply the negative against each term inside the parentheses.

### The Distributive Property of Multiplication

For all real numbers a, b, and c, $a(b+c)=ab+ac$. What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the steps we have already practiced to isolate the variable and solve the equation.

### Example

Solve for $a$. $4\left(2a+3\right)=28$

Answer: Apply the distributive property to expand $4\left(2a+3\right)$ to $8a+12$

$\begin{array}{r}4\left(2a+3\right)=28\\ 8a+12=28\end{array}$

Subtract $12$ from both sides to isolate the variable term.

$\begin{array}{r}8a+12\,\,\,=\,\,\,28\\ \underline{-12\,\,\,\,\,\,-12}\\ 8a\,\,\,=\,\,\,16\end{array}$

Divide both terms by $8$ to get a coefficient of $1$.

$\begin{array}{r}\underline{8a}=\underline{16}\\8\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\\a\,=\,\,2\end{array}$

In the video that follows, we show another example of how to use the distributive property to solve a multi-step linear equation. https://youtu.be/aQOkD8L57V0 In the next example, there are parentheses on both sides of the equal sign. To clear them, you'll need to use the distributive property on both sides of the equation.

### Example

Solve for $t$.  $2\left(4t-5\right)=-3\left(2t+1\right)$

Answer: Apply the distributive property to expand $2\left(4t-5\right)$ to $8t-10$ and $-3\left(2t+1\right)$ to$-6t-3$. Be careful in this step—you are distributing a negative number, so keep track of the sign of each number after you multiply.

$\begin{array}{r}2\left(4t-5\right)=-3\left(2t+1\right)\,\,\,\,\,\, \\ 8t-10=-6t-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$

Add $6t$ to both sides to begin combining like terms.

$\begin{array}{r}8t-10=-6t-3\\ \underline{+6t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+6t}\,\,\,\,\,\,\,\\ 14t-10=\,\,\,\,-3\,\,\,\,\,\,\,\end{array}$

Add 10 to both sides of the equation to isolate t.

$\begin{array}{r}14t-10=-3\\ \underline{+10\,\,\,+10}\\ 14t=\,\,\,7\,\end{array}$

The last step is to divide both sides by 14 to completely isolate t.

$\begin{array}{r}\Large\frac{14t}{14}\normalsize=\Large\frac{7}{14}\end{array}$

We simplify the fraction $\Large\frac{7}{14}$ into the final answer of $t=\Large\frac{1}{2}$

Watch the following video for a demonstration of how to solve a multi-step equation with two sets of parentheses. https://youtu.be/StomYTb7Xb8

## Clearing Fractions and Decimals from Equations

Sometimes, you will encounter a multi-step equation containing fractions. Before attempting to solve the equation, first use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. This will clear all the fractions out of the equation. See the example below.

### Example

Solve  $\Large\frac{1}{2}\normalsize x-3=2-\Large\frac{3}{4}\normalsize x$ by clearing the fractions in the equation first.

Answer: Multiply both sides of the equation by $4$, the common denominator of the fractional coefficients.

$\begin{array}{r}\Large\frac{1}{2}\normalsize{x-3=2-}\Large\frac{3}{4}\normalsize{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\ 4\left(\Large\frac{1}{2}\normalsize{x-3}\right)=4\left(2-\Large\frac{3}{4}\normalsize{x}\right)\end{array}$

Use the distributive property to expand the expressions on both sides. Multiply.

$\begin{array}{r}4\left(\Large\frac{1}{2}\normalsize{x}\right)-4\left(3\right)=4\left(2\right)-4\left(-\Large\frac{3}{4}\normalsize{x}\right)\\\\ \Large\frac{4}{2}\normalsize{x}-12=8-\Large\frac{12}{4}\normalsize{x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\\\ 2x-12=8-3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \end{array}$

Add 3x to both sides to move the variable terms to only one side.

$\begin{array}{r}2x-12=8-3x\, \\\underline{+3x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+3x}\\ 5x-12=8\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$

Add 12 to both sides to move the constant terms to the other side.

$\begin{array}{r}5x-12=8\,\,\\ \underline{\,\,\,\,\,\,+12\,+12} \\5x=20\end{array}$

Divide to isolate the variable.

$\begin{array}{r}\underline{5x}=\underline{20}\\ 5\,\,\,\,\,\,\,\,\,5\,\,\\ x=4\end{array}$

It is acceptable to simply do the operations on the fractions without clearing them first, but the technique used here will apply to more complicated situations you'll encounter later, so it is worthwhile to practice it. Watch the following video for a demonstration of how to solve a multi-step equation containing fractions by using the least common denominator to clear the fractions first. https://youtu.be/AvJTPeACTY0 Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find the easiest. Remember to check your answer by substituting your solution into the original equation. Sometimes, you will encounter a multi-step equation with decimals. To clear the decimals from the equation, use the multiplication property of equality to multiply both sides of the equation by a a factor of $10$ that will help clear the decimals. See the example below.

### Example

Solve $3y+10.5=6.5+2.5y$ by clearing the decimals in the equation first.

Answer: Since the smallest decimal place value represented in the equation is 0.10, we want to multiply by 10 to clear the decimals from the equation.

$\begin{array}{r}3y+10.5=6.5+2.5y\,\,\,\,\,\,\,\,\,\,\,\,\\\\ 10\left(3y+10.5\right)=10\left(6.5+2.5y\right)\end{array}$

Use the distributive property to expand the expressions on both sides.

$\begin{array}{r}10\left(3y\right)+10\left(10.5\right)=10\left(6.5\right)+10\left(2.5y\right)\end{array}$

Multiply.

$30y+105=65+25y$

Move the smaller variable term, $25y$, by subtracting it from both sides.

$\begin{array}{r}30y+105=65+25y\,\,\\ \underline{-25y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-25y} \\5y+105=65\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$

Subtract 105 from both sides to isolate the term with the variable.

$\begin{array}{r}5y+105=65\,\,\,\\ \underline{\,\,\,\,\,\,-105\,-105} \\5y=-40\end{array}$

Divide both sides by 5 to isolate the y.

$\begin{array}{r}\underline{5y}=\underline{-40}\\ 5\,\,\,\,\,\,\,\,\,\,\,\,\,5\,\,\\ y=-8\,\,\end{array}$

Watch the following example to see how to clear decimals first to solve a multi-step linear equation containing decimals. [embed]https://youtu.be/wtwepTZZnlY[/embed] Here are some steps to follow when you solve multi-step equations.

### Solving Multi-Step Equations

1. (Optional) Multiply to clear any fractions or decimals. 2. Simplify each side by clearing parentheses and combining like terms. 3. Add or subtract to isolate the variable term—you may have to move a term with the variable. 4. Multiply or divide to isolate the variable. 5. Check the solution.
Complex, multi-step equations often require multi-step solutions. Before you can begin to isolate a variable, you may need to simplify the equation first. This may mean using the distributive property to remove parentheses or multiplying both sides of an equation by a common denominator to get rid of fractions. Sometimes it requires both techniques. If your multi-step equation has an absolute value, you will need to solve two equations, sometimes isolating the absolute value expression first.