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Solving Variables on Both Sides of the Equation
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Introduction

This lesson, you will learn how to solve an equation that has variables on both sides of the equal sign. You will learn how to combine like terms, distribute when needed, add an additive inverse, and multiply by the multiplicative inverse.


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


Solving for a Variable on Both Sides of an Equation

There are equations that have variables in more than one place. For example, \( 3 + 5p = 2p - 5\).

Solving equations with variables on both sides can feel daunting, but once you learn how to use certain tools, it will feel easier. Here’s a list of some of the tools that can help you when you’re solving for a variable that appears on both sides of an equation:

  • Combine Like Terms.
    • Like terms are terms that both have the same variable with the same coefficients in them, like:
      • \(2x\) and \(5x\)
      • \(4\) and \(22.6\)
      • \(4x^{2}\) and \(7x^{2}\)
  • Distribute when needed.
    • The distributive property of multiplication is a helpful tool for simplifying equations with variables. If something within parentheses is being multiplied, you can multiply each number or variable inside the parentheses by whatever you are multiplying by to remove the parentheses:

      \begin{align*} 3(1+x) &= 3\cdot1 + 3 \cdot x &\color{red}\small\text{Distribute the \(3\) inside the parentheses} \end{align*}

      The equation is now \(3+3x\).

  • Add the additive inverse of terms to both sides.
    • The additive inverse is the number that you can add to something to make it equal 0:

\begin{align*}3x + 1 &= 5 &\color{red}\small\text{Solve for x}\\\\3x + 1 \color{red}\mathbf{-1} &=5\color{red}\mathbf{-1} &\color{red}\small\text{Additive inverse terms of \(+1\) is \(-1\)}\\\\ 3x &= 4 &\color{red}\small\text{Simplify each side} \end{align*}

  • Multiply both sides by the multiplicative inverse.
    • The multiplicative inverse is a number that you can multiply both sides of the equation by to cancel out a number that you are using to multiply. This is especially helpful if you need to isolate a variable that is part of a multiplication operation:

\begin{align*}3x&=2 &\color{red}\small\text{Solve for x}\\\\ 3x\color{red}\mathbf{(\frac{1}{3})} &= (2)\color{red}\mathbf{(\frac{1}{3})}&\color{red}\small\text{Multiplicative inverse of 3 is \(\frac{1}{3}\)}\\\\ x &= \frac{2}{3} &\color{red}\small\text{Simplify each side} \end{align*}
Remember, you can add, subtract, multiply, or divide all you want, as long as you do it to both sides of the equation.

Example 1
\begin{align*}3 + 5p &=2p -4 &\color{red}\small\text{Solve for p}\\\\3 + 5p\color{red}\mathbf{-2p} &=2p -4\color{red}\mathbf{-2p} &\color{red}\small\text{Additive inverse of 2p is -2p}\\\\3 + 3p &= -4 &\color{red}\small\text{Simplify each side}\\\\3 + 3p\color{red}\mathbf{-3} &= -4\color{red}\mathbf{-3} &\color{red}\small\text{Additive inverse of 3 is -3}\\\\3p &= -7&\color{red}\small\text{Simplify each side}\\\\3p\color{red}\mathbf{(\frac{1}{3})} &= -7\color{red}\mathbf{(\frac{1}{3})}&\color{red}\small\text{Multiplicative inverse of \(3\) is \(\frac{1}{3}\)}\\\\ p &= \frac{-7}{3} &\color{red}\small\text{Simplify each side}\\\end{align*}

The solution is \(p=-\frac{7}{3}\). In order to check if that answer is correct, solve the original equation by using \(-\frac{7}{3}\) anywhere p is used. If you do, you get the same answer on both sides of the equal sign, so the solution is correct.

Example 2
\begin{align*}7 + x + 3 &= 20 - 5x + 2x &\color{red}\small\text{Solve for \(x\)}\\\\10 + x &= 20 - 3x &\color{red}\small\text{Combine like terms}\\\\10 + x \color{red}\mathbf{-10} &=20 -3x\color{red}\mathbf{-10} &\color{red}\small\text{Additive inverse of 10 is -10}\\\\x &= 10 - 3x &\color{red}\small\text{Simplify each side}\\\\ x\color{red}\mathbf{+3x} &= 10 - 3x\color{red}\mathbf{+3x} &\color{red}\small\text{Additive inverse of -3x is 3x}\\\\4x &= 10&\color{red}\small\text{Simplify each side}\\\\4x\color{red}\mathbf{(\frac{1}{4})} &= 10\color{red}\mathbf{(\frac{1}{4})}&\color{red}\small\text{Multiplicative inverse of 4 is \(\frac{1}{4}\) }\\\\x &= \frac{10}{4}&\color{red}\small\text{Simplify each side}\\\\x &= \frac{5}{2}&\color{red}\small\text{Simplify the fraction} \end{align*}

Initially, you got \(x=\frac{10}{4}\), but because 10 and 4 have common factors, you can simplify this fraction. 10 can be factored into \(5\times2\), and 4 can be factored into \(2\times2\). Because both numbers have a factor of 2, you can cancel out a 2 from each of the factorizations. The final answer is \(x=\frac{5}{2}\).


Things to Remember


  • You can perform any operation that you need to when solving for a variable as long as you perform it on both sides of the equation.
  • When solving an equation with variables on both sides, move all of the variables to one side and all of the regular numbers to the opposite side through combining like-terms.
  • Remember to use tools like combining like-terms, distributing when needed, adding an additive inverse, and multiplying by the multiplicative inverse to simplify the equation.

Practice Problems

Solve for the variable:
  1. \(2 - 7{\text{g}} = -9{\text{g}}\) (
    Solution
    x
    Solution: \(-1\)
    )
  2. \(12 + 3{\text{W}} = -4 + {\text{W}}\) (
    Solution
    x
    Solution: \(-8\)
    Details:
    Do the order of operations backwards to solve for W.

    Step 1: Combine like terms using addition or subtraction. Right now you have terms containing W on both sides of the equation:

    \(12 + 3{\text{W}} = -4 + {\text{W}}\)

    First, subtract W from both sides to gather all terms containing W to the left side of the equation:

    This image shows the equation twelve plus three w equals negative four plus w. Under the w on the left-hand side is a red negative w and under the w on the right-hand side is another red negative w.

    Which gives you:

    \(12 + 2{\text{W}} = -4 {\color{Red} + 0}\)

    Which is equal to:

    \(12 + 2{\text{W}} = -4\)

    Next, subtract 12 from both sides of the equation:

    The image shows the equation twelve plus two w equals negative four. Under twelve on the left-hand side of the equal sign is a red negative twelve. Under the negative four on the right-hand side is another red negative twelve.

    Which gives you:

    \(2{\text{W}} = {\color{Red} - 16}\)

    Step 2: Undo any multiplication using the multiplicative inverse or division to isolate W.

    \({\color{Cyan} \left [ \dfrac{1}{2} \right ]}2{\text{W}}=-16{\color{Cyan} \left [ \dfrac{1}{2} \right ]}\)

    On the left side of the equation: \(\left (\dfrac{1}{2} \right )\left ( 2 \right )\) gives you 1W.

    On the right side of the equation: \(-16\dfrac{1}{2}=\dfrac{-16}{1}\times\dfrac{1}{2}=\dfrac{-16\times1}{1\times2}=\dfrac{-16}{2}=-8\).

    The final answer is: \({\text{W}} = {\color{Red} - 8}\).
    )
  3. \({\text{m}} {-} 3 = 2{\text{m}} - 3\) ( )
  4. \(3 - 6{\text{P}} = -6 - 7{\text{P}}\) (
    Solution
    x
    Solution: \(-9\)
    Details:
    Do the order of operations backwards to solve for W.

    Step 1: Combine like terms using addition or subtraction.

    Right now you have terms containing P on both sides of the equation:

    \(3 - 6{\text{P}} = -6 - 7{\text{P}}\)

    First, add 7P to both sides to gather all terms containing P to the left side of the equation:

    This image shows the equation three minus six p equals negative six minus seven p. Below the minus six p on the left-hand side is a red positive seven p. Below the minus seven p on the right-hand side is another red positive seven p.

    Which gives you:

    The image shows two equations. The top equation is three plus p equals negative six plus zero. The p on the left-hand side is red and the plus zero on the right-hand side is also red, indicating they are the result of the previous operations. Below the equal sign is an arrow pointing down to the bottom equation. The bottom equation is three plus p equals negative six.

    Next, subtract 3 from both sides of the equation:

    This image shows the bottom equation from the previous images. Three plus p equals negative six. Under the three on the left-hand side is a red negative three. Under the negative six on the right-hand side is another red negative three.

    Which gives you:

    \({\text{P}} = {\color{Red} -9}\)
    )
  5. \(6{\text{x}} {-} 1 = -5 + 7{\text{x}}\) (
    Solution
    x
    Solution: 4
    )
  6. \(7 - 5{\text{C}} = -9 - 9{\text{C}}\) ( )

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