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## How to Solve Equations with Variables on Both Sides

Last Updated: March 11, 2023 Fact Checked

This article was co-authored by JohnK Wright V . JohnK Wright V is a Certified Math Teacher at Bridge Builder Academy in Plano, Texas. With over 20 years of teaching experience, he is a Texas SBEC Certified 8-12 Mathematics Teacher. He has taught in six different schools and has taught pre-algebra, algebra 1, geometry, algebra 2, pre-calculus, statistics, math reasoning, and math models with applications. He was a Mathematics Major at Southeastern Louisiana and he has a Bachelor of Science from The University of the State of New York (now Excelsior University) and a Master of Science in Computer Information Systems from Boston University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 180,686 times.

To study algebra, you will see equations that have a variable on one side, but later on you will often see equations that have variables on both sides. The most important thing to remember when solving such equations is that whatever you do to one side of the equation, you must do to the other side. Using this rule, it is easy to move variables around so that you can isolate them and use basic operations to find their value.

## Solving Equations with One Variable on Both Sides

## Solving System Equations with Two Variables

## Solving Example Problems

## Community Q&A

## Things You'll Need

You might also like.

- ↑ https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-distributive-property/v/the-distributive-property
- ↑ https://www.virtualnerd.com/algebra-1/linear-equations-solve/variables-both-sides-equations/variables-both-sides-solution/variables-grouping-symbols-both-sides
- ↑ https://www.youtube.com/watch?v=hrAOSknrYiI&t=296s
- ↑ https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-expressions-and-variables/cc-6th-evaluating-expressions/v/expression-terms-factors-and-coefficients
- ↑ https://www.virtualnerd.com/pre-algebra/linear-functions-graphing/system-of-equations/solving-systems-equations/two-equations-two-variables-substitution

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## Solving for a Variable:

Solving variables on both sides of the equation.

So far we’ve only seen equations with a single variable. There are equations that have variables in more than one place. For example, \(3{\text{x}} + 4 = {\text{x}}\). How do we solve these? The first video will explain some of the tools we will use, then the second video will show how to solve these kinds of equations.

Video Source (03:22 mins) | Transcript

Video Source (09:43 mins) | Transcript

Tools taught in the first video:

- Combine like Terms (add things that have the same variable)
- Distribute when needed (multiply each of the things inside the parentheses)
- Add the additive inverse of terms to both sides
- Multiply by the multiplicative inverse to both sides

When faced with a problem, start by combining any like terms on the same side of the equation. Then combine like terms from both sides of the equation. After that, use the things we learned in last week’s lesson of adding or multiplying by the inverse as needed. Remember, we can add, subtract, multiply, or divide all we want, as long as we do it to both sides of the equation.

## Additional Resources

- Khan Academy: Combining Like Terms (03:05 mins, Transcript )
- Khan Academy: Introduction to Equations with Variables on Both Sides (08:52 mins, Transcript )
- Khan Academy: Linear Equations 3 (06:44 mins, Transcript )

Practice Problems

Solve for the following variables:

- \(2 - 7{\text{g}} = -9{\text{g}}\)

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

- \({\text{m}} {-} 3 = 2{\text{m}} - 3\)

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

- \(6{\text{x}} {-} 1 = -5 + 7{\text{x}}\)
- \(7 - 5{\text{C}} = -9 - 9{\text{C}}\)

We do the order of operations backwards to solve for \({\text{W}}\).

Step 1: Combine like terms using addition or subtraction. Right now we have terms containing \({\text{W}}\) on both sides of the equation:

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

Which gives us:

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

Which is equal to:

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

Next, we’ll subtract 12 from both sides of the equation:

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

Step 2: Undo any multiplication using the multiplicative inverse or division to isolate \({\text{W}}\).

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

On the left side of the equation: \(\left (\frac{1}{2} \right )\left ( 2 \right )\) gives us \(1{\text{W}}\).

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

So our final answer is:

\({\text{W}} = {\color{Red} - 8}\)

- 0 ( Solution Video | Transcript )

Step 1: Combine like terms using addition or subtraction.

Right now we have terms containing \({\text{P}}\) on both sides of the equation:

First, we’ll add \(7{\text{P}}\) to both sides to gather all terms containing \({\text{P}}\) to the left side of the equation:

Next, we’ll subtract 3 from both sides of the equation:

\({\text{P}} = {\color{Red} -9}\)

- \(-4\) ( Solution Video | Transcript )

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