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Solving for a Variable Using Addition, Subtraction, Multiplication, and Division with Fractions
> ... Math > Solving for a Variable > 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 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{navy}\small\text{Solve for \(x\)}\\\\\frac{3}{8}x - \frac{1}{2} \color{green}\mathbf{+\frac{1}{2}} &= \frac{1}{4}\color{green}\mathbf{+\frac{1}{2}} &\color{navy}\small\text{Additive inverse of \(-\frac{1}{2}\) is \(+\frac{1}{2}\)}\\\\ \frac{3}{8}x &= \frac{1}{4}\color{green}\mathbf{+\frac{1}{2}} &\color{navy}\small\text{\(-\frac{1}{2}\) and \(+\frac{1}{2}\) cancel out}\\\\ \frac{3}{8}x & = \frac{1}{4} + \frac{2}{4} &\color{navy}\small\text{Find Common Denominator}\\\\\frac{3}{8}x & = \frac{3}{4} &\color{navy}\small\text{Add}\\\\\frac{3}{8}x \color{green}\mathbf{(\frac{8}{3})} &= \frac{3}{4}\color{green}\mathbf{(\frac{8}{3})} &\color{navy}\small\text{Multiplicative inverse of \(\frac{3}{8}\) is \(\frac{8}{3}\)}\\\\\frac{24}{24}x & = \frac{24}{12} &\color{navy}\small\text{Multiply}\\\\x & = 2 &\color{navy}\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{navy}\small\text{Solve for x} \\\\3(\frac{-2x + 8}{5} - 3) + 17 \color{green}\mathbf{-17} &= 20 \color{green}\mathbf{-17} & \color{navy}\small\text{Additive inverse of \(+17\) is \(-17\)} \\\\3(\frac{-2x + 8}{5} - 3) &= 3 & \color{navy}\small\text{Simplify each side} \\\\\frac{\cancel 3(\frac{-2x + 8}{5} - 3)}{ \color{green}\mathbf{\cancel 3}} &= \frac{\cancel 3}{ \color{green}\mathbf{{\cancel 3}}} & \color{navy}\small\text{Divide each side by 3 & cancel the 3's} \\\\\frac{-2x + 8}{5} - 3 &= 1 & \color{navy}\small\text{Simplify each side} \\\\\frac{-2x + 8}{5} - 3 \color{green}\mathbf{+3} &= 1 \color{green}\mathbf{+3} & \color{navy}\small \text{Additive inverse of \(-3\) is \(+3\)} \\\\\frac{-2x + 8}{5} &= 4 & \color{navy}\small\text{Simplify each side} \\\\\frac{-2x + 8}{5} \color{green}\mathbf{(5)} &= 4 \color{green}\mathbf{(5)} & \color{navy}\small \text{Multiplicative inverse of \(\frac{1}{5}\) is \(5\)} \\\\ \frac{-2x + 8}{\cancel 5} \color{green}\mathbf{(\cancel 5)} &= 4 \color{green}\mathbf{(5)} & \color{navy}\small \text{\(\frac{1}{5}\) and \(5\) cancel out} \\\\-2x + 8 &= 20 & \color{navy}\small\text{Simplify each side} \\\\-2x + 8 \color{green}\mathbf{-8} &= 20 \color{green}\mathbf{-8} & \color{navy}\small\text{Additive inverse of \(+8\) is \(-8\)} \\\\-2x &= 12 & \color{navy}\small\text{Simplify each side} \\\\-2x \color{green}\mathbf{(\frac{1}{-2})} &= 12 \color{green}\mathbf{(\frac{1}{-2})} & \color{navy}\small\text{Multiplicative inverse of \(-2\) is \(\frac{1}{-2}\)} \\\\\cancel {-2}x \color{green}\mathbf{(\frac{1}{\cancel {-2}})} &= \frac{12}{-2}& \color{navy}\small\text{Cancel the \(-2\)'s} \\\\ x &= \frac{12}{-2}& \color{navy}\small\text{The value of \(x\) is isolated} \\\\ x &= -6 & \color{navy}\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

Solve for the variable:
  1. \(\dfrac{3}{2}{\text{x}} + \dfrac{1}{4} = \dfrac{13}{4}\) (
    Solution
    x
    Solution: 2
    )
  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 dotted line running vertically through the equal signs, separating the left and right sides of the equations. The first line is the equation one-third minus two-thirds x equals 3. The second line is the equation one-third plus negative two-thirds x equals 3. 

    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 \({\color{navy}{\text{add a negative}}}\) \(\dfrac{1}{3}\) to each side.

    This image is divided into two sides with a dotted line running vertically through the equal sign, separating the left and right sides of the equation. The equation is one-third plus negative one-third plus negative two-thirds x equals 3 plus negative one-third. The negative one-third on both sides of the equation are both in red, indicating that these are new to the equation. 

    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.

    This image is divided into two sides with a dotted line running vertically through the equal signs, separating the left and right sides of three equations. The first equation is one-third plus negative one-third plus negative two-thirds x equals 3 plus negative one-third. The second equation has a bracket showing that the first part of the first equation becomes 0. The second equation is zero plus negative two-thirds x equals 3 plus negative one-third. The third equation is negative two-thirds x equals 3 plus negative one-third. 

    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 dotted line running vertically through the equal signs, separating the left and right sides of three equations. The first equation is negative two-thirds x equals 3 plus negative one-third. The second equation is negative two-thirds x equals 3 over 1 times three-thirds plus negative one-third. The third equation has a bracket showing that the 3 over 1 times three-thirds in the second equation becomes nine-thirds in the third equation. The third equation is negative two-thirds x equals nine-thirds plus negative one-third. 

    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 dotted line running vertically through the equal signs, separating the left and right sides of three equations. The first equation is negative two-thirds x equals nine-thirds plus negative one-third. The second equation is negative two-thirds x equals 9 minus 1 over 3. The third equation is negative two-thirds x equals 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 dotted line running vertically through the equal signs, separating the left and right sides of three equations. The first equation is negative three-halves times negative two-thirds x equals negative three-halves times eight-thirds. The second equation is one x equals negative 24-sixths. The third equation is x equals negative 4. 

    \(-\dfrac{24}{6}=-4\)

    The final solution is: \({\text{x}} = -4\).
    )
  3. \(3=-2{\text{D}}+\dfrac{2}{3}\) (
    Solution
    x
    Solution: \(-\dfrac{7}{6}\)
    )
  4. \(-\dfrac{2}{3}=\dfrac{2}{3}{\text{U}}{-}\dfrac{1}{2}\) (
    Video Solution
    x
    Solution: \(-\dfrac{1}{4}\)
    Details:

     | Transcript)
  5. \(8 = (5{\text{x}} {-} 2) - 5\) (
    Solution
    x
    Solution: 3
    )
  6. \(2(\dfrac{{\text{F}}+1}{-2})-8=-4\) (
    Video Solution
    x
    Solution: \(-5\)
    Details:

     | Transcript)
  7. \(1.3{\text{d}} + 5.2 = -2.6\) (
    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.

    Two equations, one on top of the other. The first equation is one-point-three d plus five-point-two equals negative two-point-six. The second equation is one-point-three d plus five-point-two plus negative five-point-two equals negative two-point-six plus negative five-point-two. The negative five-point-two on both sides of the equal sign are gold to indicate these are new. 

    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.

    Three equations, one on top of the other. The first equation is one-point-three d plus five-point-two plus negative five-point-two equals negative two-point-six plus negative five-point-two. There is a bracket showing that the five-point-two plus negative five-point-two in the first equation becomes zero in the second equation. There is another bracket showing that the negative two-point-six plus negative five-point-two in the first equation becomes negative seven-point-eight in the second equation. The second equation is one-point-three d plus zero equals negative seven-point-eight. The third equation is one-point-three d equals negative seven-point-eight. 

    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 \(\dfrac{1}{1.3}\)

    Two equations, one above the other. The first equation is one-point-three d equals (negative seven-point-eight). The second equation is the same as the first except 1 over 1.3 is multiplied by each side of the equation. The second equation is (1 over 1.3) times 1.3 d equals (1 over 1.3) 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.

    Two equations, one over the other. The first equation is the same as the bottom equation of the previous image, which is (one-over-one-point-three) times one-point-three d equals (one-over-one-point-three) times (negative seven-point-eight). The fractions (one-over-one-point-three) on each side of the equal sign are blue. The second equation is (one-point-three over one-point-three) d equals (negative seven-point-eight over one-point-three). On both sides of the equation, the one-point-three is blue to help identify where it came from in the previous equation. 

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

    Be sure not to lose the negative in the final solution.
    )
Solving for a Variable on One Side Using Multiplication and Division with Fractions
Solving Variables on Both Sides of the Equation