Solving for a Variable Using Addition, Subtraction, Multiplication, and Division with Fractions

Introduction

In this lesson, you will practice solving equations for a variable when fractions are involved.

These videos illustrate the lesson material below. Watching the videos is optional.

Solving for a Variable on One Side: Part 6 - Add, Subtract, Multiply, or Divide with fractions (06:53 mins) |Transcript
Examples of Solving for a Variable on One Side with Many Operations Involved (12:45 mins) |Transcript

Solving Multi-Step Equations for a Variable on One Side

Sometimes solving for a variable requires more than one step. This lesson demonstrates how to solve for a variable when there is addition or subtraction as well as multiplication involving fractions. It is important to understand that the concepts don’t change whether there are whole numbers, decimals, or fractions in the equation. The principles of how to solve for a variable are still the same.

Example 1
Solve for $x$: $\frac{3}{8}x - \frac{1}{2} = \frac{1}{4}$

The goal of these kinds of problems is to isolate $x$. Do that by working backwards through the order of operations. In this equation, that means you begin by adding $\frac{1}{2}$ and then multiplying by the multiplicative inverse to both sides.

\begin{align*}\frac{3}{8}x - \frac{1}{2} & = \frac{1}{4} &\color{red}\small\text{Solve for $x$}\\\\\frac{3}{8}x - \frac{1}{2} \color{red}\mathbf{+\frac{1}{2}} &= \frac{1}{4}\color{red}\mathbf{+\frac{1}{2}} &\color{red}\small\text{Additive inverse of $-\frac{1}{2}$ is $+\frac{1}{2}$}\\\\ \frac{3}{8}x &= \frac{1}{4}\color{red}\mathbf{+\frac{1}{2}} &\color{red}\small\text{$-\frac{1}{2}$ and $+\frac{1}{2}$ cancel out}\\\\ \frac{3}{8}x & = \frac{1}{4} + \frac{2}{4} &\color{red}\small\text{Find Common Denominator}\\\\\frac{3}{8}x & = \frac{3}{4} &\color{red}\small\text{Add}\\\\\frac{3}{8}x \color{red}\mathbf{(\frac{8}{3})} &= \frac{3}{4}\color{red}\mathbf{(\frac{8}{3})} &\color{red}\small\text{Multiplicative inverse of $\frac{3}{8}$ is $\frac{8}{3}$}\\\\\frac{24}{24}x & = \frac{24}{12} &\color{red}\small\text{Multiply}\\\\x & = 2 &\color{red}\small\text{Simplify}\\\end{align*}

Example 2
Solve for $x$: \begin{align*} 3(\frac{-2x + 8}{5} - 3) + 17 = 20 \end{align*}

Break down these large problems into steps using the Order of Operations (PEMDAS), then undo these steps by going backward and doing the inverse operation to find out what the variable is. It is important that you do each of these operations to both sides of the equation.

\begin{align*}3(\frac{-2x + 8}{5} - 3) + 17 &= 20 & \color{red}\small\text{Solve for x} \\\\3(\frac{-2x + 8}{5} - 3) + 17 \color{red}\mathbf{-17} &= 20 \color{red}\mathbf{-17} & \color{red}\small\text{Additive inverse of $+17$ is $-17$} \\\\3(\frac{-2x + 8}{5} - 3) &= 3 & \color{red}\small\text{Simplify each side} \\\\\frac{\cancel 3(\frac{-2x + 8}{5} - 3)}{ \color{red}\mathbf{\cancel 3}} &= \frac{\cancel 3}{ \color{red}\mathbf{{\cancel 3}}} & \color{red}\small\text{Divide each side by 3 & cancel the 3's} \\\\\frac{-2x + 8}{5} - 3 &= 1 & \color{red}\small\text{Simplify each side} \\\\\frac{-2x + 8}{5} - 3 \color{red}\mathbf{+3} &= 1 \color{red}\mathbf{+3} & \color{red}\small \text{Additive inverse of $-3$ is $+3$} \\\\\frac{-2x + 8}{5} &= 4 & \color{red}\small\text{Simplify each side} \\\\\frac{-2x + 8}{5} \color{red}\mathbf{(5)} &= 4 \color{red}\mathbf{(5)} & \color{red}\small \text{Multiplicative inverse of $\frac{1}{5}$ is $5$} \\\\ \frac{-2x + 8}{\cancel 5} \color{red}\mathbf{(\cancel 5)} &= 4 \color{red}\mathbf{(5)} & \color{red}\small \text{$\frac{1}{5}$ and $5$ cancel out} \\\\-2x + 8 &= 20 & \color{red}\small\text{Simplify each side} \\\\-2x + 8 \color{red}\mathbf{-8} &= 20 \color{red}\mathbf{-8} & \color{red}\small\text{Additive inverse of $+8$ is $-8$} \\\\-2x &= 12 & \color{red}\small\text{Simplify each side} \\\\-2x \color{red}\mathbf{(\frac{1}{-2})} &= 12 \color{red}\mathbf{(\frac{1}{-2})} & \color{red}\small\text{Multiplicative inverse of $-2$ is $\frac{1}{-2}$} \\\\\cancel {-2}x \color{red}\mathbf{(\frac{1}{\cancel {-2}})} &= \frac{12}{-2}& \color{red}\small\text{Cancel the $-2$'s} \\\\ x &= \frac{12}{-2}& \color{red}\small\text{The value of $x$ is isolated} \\\\ x &= -6 & \color{red}\small\text{Simplify} \end{align*}

Things to Remember

Rewrite the problem after completing each step to prevent losing track of what has already been done.
A fraction is a form of division.
When isolating a variable, remember to perform the opposite operations on both sides of the equation in order to cancel them out.

Practice Problems

$\dfrac{3}{2}{\text{x}} + \dfrac{1}{4} = \dfrac{13}{4}$ (
Solution

x

Solution: 2

)
$\dfrac{1}{3} - \dfrac{2}{3}{\text{x}} = 3$ (
Solution
x

Solution: $-4$
Details:
One thing you may find helpful is to first change any subtraction to the addition of a negative number. This can help avoid losing the negative.

$This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of two equations, one above the other. The first equation on the left-hand side shows the expression one-third minus two-thirds x. The minus or subtraction symbol is in red, bringing it to our attention. There is an equal sign to the right of this expression along the light gray line representing that the expression is equal to whatever is on the right-hand side. On the right-hand side of this equal sign is the number 3. Below this equation, the same equation has been written again, but with some variations. Instead of the minus sign, there is now a plus sign in green and the fraction negative two-thirds is in parentheses and the negative is red, showing it is the same as the subtraction sign from the previous equation. So the second expression on the left-hand side reads one-third plus (negative two-thirds)x. The negative two-thirds is all multiplied to the x. The equal sign is in the middle along the gray line, and to the right on the right-hand side is the number 3 again.$

To solve for the variable x, use the order of operations backwards, but first, find the steps going forward through the order of operations assuming you had a specific number for x.

If you were going forward, the steps would go as follows:
Parentheses: None this time
Exponents: None this time
Multiplication & Division: between the $\left (-\dfrac{2}{3} \right )$ and x, so first multiply the $\left (-\dfrac{2}{3} \right ){\text{x}}$
Addition & Subtraction: Add $\dfrac{1}{3}$ to the previous answer.

Going backwards to solve for a variable:

Since you are solving for a variable, do these steps in reverse.

Step 1: Do the reverse of adding $\dfrac{1}{3}$ to each side of the equation. In other words, subtract $\dfrac{1}{3}$ from each side or ${\color{Red}{\text{add a negative}}}$ $\dfrac{1}{3}$ to each side.

On the left-hand side:

The additive inverses $-\dfrac{1}{3}$ and $+\dfrac{1}{3}$ add together to 0. This leaves just $\left (-\dfrac{2}{3} \right ){\text{x}}$ on the left side.

On the right-hand side:

You have $3+\left (-\dfrac{1}{3} \right )$.

Adding fractions requires common denominators. The number 3 is a whole number, so it has an invisible denominator of 1. You can rewrite it as $\dfrac{3}{1}$

The greatest common denominator between 1 and 3 is 3. Multiply $\dfrac{3}{1}$ by $\dfrac{3}{3}$ to get the common denominator 3.

$This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of three equations, one above the other. The first equation on the left-hand side shows the expression negative two-thirds x. The left-hand side expressions all stay the same in this image, but the right-hand side changes, showing the steps for solving the right-hand side. The top expression on the right-hand side is three plus (negative one-third). The three is in blue indicating that something is going to happen to it in the next step. The next line on the right-hand side of the equation shows is (three-over-one) times three-over-three plus (negative one-third). The three-over-one is blue showing it is the same as the three in the previous equation. The three-over-three is green to emphasize that it is new. Below the (three-over-one) times three-over-three is a green bracket pointing down to the next version of the equation. Just below the bracket is the fraction nine-over-three. This is added to (negative one-third). The fraction nine-over-three is also green to represent that it is the product of the blue three-over-one and the green three-over-three.$

When adding or subtracting fractions with common denominators, simply add or subtract the numbers in the numerator. The denominator stays the same because that is the part of the fraction that tells you what sizes the pieces are.

$\left ( \dfrac{9}{3} \right )+\left ( -\dfrac{1}{3} \right )=\dfrac{8}{3}$

Subtract $9-1$. The negative on the $\left ( \dfrac{1}{3} \right )$ is the same as adding a negative 1 or subtracting 1.

This leaves you with $\left ( -\dfrac{2}{3} \right ){\text{x}}$ on the left and $\dfrac{8}{3}$ on the right.

$This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of three equations, one above the other. The first equation on the left-hand side shows the expression negative two-thirds x. It is equal to the expression on the right-hand side that says nine-thirds plus (negative one-third). The nine-thirds is in green to show we are working with it. The negative on the one-third is in red to highlight that it is negative. The next equation down also has negative two-thirds x on the left-hand side. It is equal to the expression on the right-hand side which shows a fraction where the numerator is a green nine and then a red subtraction symbol followed by a one. The green nine comes from the numerator of the nine-thirds in the previous equation. The red subtraction symbol comes from the negative symbol in the previous equation. The one comes from the one-third in the previous equation. The denominator of this fraction is 3. The last equation at the bottom of this image has negative two-thirds x on the left-hand side. To the right of this is the equal sign and the fraction eight-thirds. The eight is colored blue to represent it is the result of nine minus 1 in the previous equation.$

Step 2: Multiply both sides by the multiplicative inverse of $-\dfrac{2}{3}$.

This will leave just 1x on the left-hand side.

$This image is divided into two sides with a very light gray dotted line running vertically through the image. At the top of the left side it says, “Left-hand side.” At the top of the right side it says, “Right-hand side.” These words are in light gray indicating they are not part of the math problem. The light gray line travels down through the equal signs of three equations, one above the other. The first equation on the left-hand side shows the expression (negative three-halves)(negative two-thirds) x. The parentheses express that the negative three-halves is being multiplied to the negative two-thirds. All of it is multiplied to x. The negative three-halves is in blue to show it is new. The negative is in red to highlight it. To the right is the equal sign. On the right-hand side of the equal sign is the expression (negative three-halves) eight-thirds. The parentheses next to the eight-thirds indicate the two fractions are being multiplied together. The negative is red to emphasize it and the three-halves is blue to show it is new. Below this equation, on the left-hand side is 1x. The equal sign is between the left and right-hand sides. On the right-hand side is the fraction negative twenty-four-sixths meaning twenty-four in the numerator and six in the denominator. The negative is red to highlight it and the fraction twenty-four-sixths is green to show it is the solution of the previous right-hand side expression. The equation at the bottom is x on the left-hand side equal to negative 4 on the right-hand side. The negative is red to highlight it and the four is green to indicate it is the solution of the previous expression.$

$-\dfrac{24}{6}=-4$

The final solution is: ${\text{x}} = -4$.
)
$3=-2{\text{D}}+\dfrac{2}{3}$ (
Solution

x

Solution: $-\dfrac{7}{6}$

)
$-\dfrac{2}{3}=\dfrac{2}{3}{\text{U}}{-}\dfrac{1}{2}$ (
Video Solution

x

Solution: $-\dfrac{1}{4}$
Details:

(Solving for a Variable Using Addition, Subtraction, Multiplication, and Division with Fractions #4 (05:47 mins) | Transcript)

| Transcript)
$8 = (5{\text{x}} {-} 2) - 5$ (
Solution

x

Solution: 3

)
$2\left ( \dfrac{{\text{F}}+1}{-2}\right )-8=-4$ (
Video Solution

x

Solution: $-5$
Details:

(Solving for a Variable Using Addition, Subtraction, Multiplication, and Division with Fractions # 6 (04:02 mins) | Transcript)

| Transcript)
$1.3{\text{d}} + 5.2 = -2.6$ (
Solution

x

Solution: $-6$
Details:
Use the order of operations backwards to unravel this equation and solve for the variable d. This example uses decimal numbers. Treat them the same as integers or fractions.

Step 1: Do the opposite of any addition or subtraction (S&A in PEMDAS).

In this example, add the additive inverse of positive 5.2 to both sides.

The additive inverse is $-5.2$, so you need to add $-5.2$ to both sides of the equation.

The left side of the equal sign:

$5.2 + -5.2 = 0$ leaving $1.3{\text{d}}$ on the left.

The right side of the equal sign:

$-2.6 + -5.2 = -7.8$

Remember the rules of adding a negative and a negative. They just become more negative.

Step 2: Do the opposite of any multiplication or division (D&M in PEMDAS).

In this example, multiply both sides by the multiplicative inverse of $1.3$. The multiplicative inverse of a decimal is found the same way as the multiplicative inverse of an integer: $\dfrac{1}{\text{decimal}}$.

Multiply both sides by $\dfrac{1}{1.3}$

$This image shows two equations, one above the other. The top equation is one-point-three d equals (negative seven-point-eight). The negative is red to bring attention to it. The bottom equation is the same as the top equation except that both sides of the equal sign have the fraction (one-over-one-point-three) in blue multiplied by what was originally there. In other words, (one-over-one-point-three) times one-point-three d equals (one-over-one-point-three) times (negative seven-point-eight).$

The left side of the equation:

$\left ( \dfrac{1}{1.3} \right )\left ( -1.3 \right ) = 1$ so you are left with 1d.

The right side of the equation:

$\left ( \dfrac{1}{1.3} \right )\left ( -7.8 \right ) = \dfrac{\left ( -7.8 \right )}{\left ( 1.3 \right )}=-6$

You can solve this using longhand or a calculator.

$This image shows two equations, one over the other. The top equation is the same as the bottom equation of the previous image. It is (one-over-one-point-three) times one-point-three d equals (one-over-one-point-three) times (negative seven-point-eight). The fraction (one-over-one-point-three) on each side of the equal sign is blue. The bottom equation on the left-hand side shows a fraction whose numerator is one-point-three and denominator is a blue one-point-three to help show where the numbers went in the multiplication. This fraction is all multiplied to the variable d. The left-hand side is equal to the fraction negative seven-point-eight over one-point-three. Again the one-point-three is blue to help identify where it came from.$

The final solution is: ${\text{d}} = -6$.

Be sure not to lose the negative in the final solution.

)

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