How to Solve Systems of Equations with Substitution?

The substitution method is one of the most reliable ways to solve a system of equations with two variables. The idea is simple: solve one equation for one variable, then plug that expression into the other equation to reduce it to a single-variable problem you already know how to solve. This guide walks through every step with worked examples and practice problems.

What Is the Substitution Method?

The substitution method involves replacing one variable in an equation with an expression from another equation. This turns a two-variable system into a one-variable equation that is easy to solve. It works best when one equation is already solved for a variable or can be rearranged with little effort.

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How to Use Substitution to Solve a System

Step 1: Solve One Equation for One Variable

If one equation is not already in the form \(\color{blue}{x = \ldots}\) or \(\color{blue}{y = \ldots}\), rearrange it.

• \(\color{blue}{x + y = 5}\)  →  \(\color{blue}{y = 5 – x}\)

Step 2: Substitute into the Other Equation

Replace the variable in the second equation with the expression you found.

• If \(\color{blue}{y = 5 – x}\), substitute into \(\color{blue}{3x + y = 11}\):
  \(\color{blue}{3x + (5 – x) = 11}\)  →  \(\color{blue}{2x + 5 = 11}\)

Step 3: Solve the Resulting Equation

You now have a single-variable equation. Solve it.

• \(\color{blue}{2x + 5 = 11}\)  →  \(\color{blue}{2x = 6}\)  →  \(\color{blue}{x = 3}\)

Step 4: Back-Substitute to Find the Other Variable

Substitute the value you found back into the expression from Step 1.

• \(\color{blue}{y = 5 – 3 = 2}\)  →  solution: \(\color{blue}{(3, 2)}\)

Step-by-Step Summary

1. Choose one equation and solve it for one variable.
2. Substitute that expression into the other equation.
3. Solve the resulting single-variable equation.
4. Substitute back to find the second variable.
5. Check your solution in both original equations.

Watch: Solving Systems by Substitution

Khan Academy demonstrates the full substitution method process:

Systems with Substitution – Worked Examples

Example 1: Solve: \(\color{blue}{y = 2x + 1}\) and \(\color{blue}{3x + y = 16}\).

Substitute \(\color{blue}{y = 2x + 1}\) into the second equation:
\(\color{blue}{3x + (2x + 1) = 16}\)  →  \(\color{blue}{5x + 1 = 16}\)  →  \(\color{blue}{5x = 15}\)  →  \(\color{blue}{x = 3}\).
Back-substitute: \(\color{blue}{y = 2(3) + 1 = 7}\).
Solution: \(\color{blue}{(3, 7)}\).

Example 2: Solve: \(\color{blue}{x = y – 3}\) and \(\color{blue}{2x + y = 12}\).

Substitute \(\color{blue}{x = y – 3}\):
\(\color{blue}{2(y – 3) + y = 12}\)  →  \(\color{blue}{3y – 6 = 12}\)  →  \(\color{blue}{3y = 18}\)  →  \(\color{blue}{y = 6}\).
Back-substitute: \(\color{blue}{x = 6 – 3 = 3}\).
Solution: \(\color{blue}{(3, 6)}\).

Example 3: Solve: \(\color{blue}{x + 2y = 10}\) and \(\color{blue}{3x – y = 9}\).

From the first equation: \(\color{blue}{x = 10 – 2y}\).
Substitute: \(\color{blue}{3(10 – 2y) – y = 9}\)  →  \(\color{blue}{30 – 7y = 9}\)  →  \(\color{blue}{7y = 21}\)  →  \(\color{blue}{y = 3}\).
Back-substitute: \(\color{blue}{x = 10 – 2(3) = 4}\).
Solution: \(\color{blue}{(4, 3)}\).

Example 4: Solve: \(\color{blue}{2x + 3y = 13}\) and \(\color{blue}{y = x – 1}\).

Substitute \(\color{blue}{y = x – 1}\):
\(\color{blue}{2x + 3(x – 1) = 13}\)  →  \(\color{blue}{5x – 3 = 13}\)  →  \(\color{blue}{x = \frac{16}{5} = 3.2}\).
Back-substitute: \(\color{blue}{y = 3.2 – 1 = 2.2}\).
Solution: \(\color{blue}{(3.2, 2.2)}\).

More Practice: The Substitution Method Explained

This additional Khan Academy video offers more examples with step-by-step narration:

Exercises for Systems of Equations with Substitution

1. Solve: \(\color{blue}{y = x + 2}\) and \(\color{blue}{x + y = 10}\)
2. Solve: \(\color{blue}{x = 3y}\) and \(\color{blue}{x + 2y = 15}\)
3. Solve: \(\color{blue}{y = -x + 6}\) and \(\color{blue}{2x + 3y = 14}\)
4. Solve: \(\color{blue}{x – y = 4}\) and \(\color{blue}{3x + y = 16}\)
5. Solve: \(\color{blue}{2x – y = 5}\) and \(\color{blue}{x + 3y = 10}\)

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Answers

1. \(\color{blue}{(4, 6)}\)
2. \(\color{blue}{(9, 3)}\)
3. \(\color{blue}{(4, 2)}\)
4. \(\color{blue}{(5, 1)}\)
5. \(\color{blue}{(\frac{25}{7}, \frac{15}{7})}\)  ≈  \(\color{blue}{(3.57, 2.14)}\)

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Free Systems of Equations with Substitution Worksheet

Ready to practice on your own? Download our free Systems of Equations with Substitution worksheet below, work through each problem at your own pace, and then check your answers. If a few give you trouble, scroll back up to the worked examples and try again — steady practice is the surest way to master Systems of Equations with Substitution before a quiz or test.

Download Solving Systems by Substitution Worksheet

Frequently Asked Questions

When is substitution better than elimination?

Substitution is most efficient when one equation is already solved for a variable, or when one variable has a coefficient of 1 \(\color{blue}{\text{ or } -1}\), making it easy to isolate. Elimination is faster when both equations are in standard form with coefficients that are easy to match.

What if both equations need to be rearranged?

Pick the simpler-looking equation and solve for the variable with the smaller coefficient. This minimizes fractions in your work.

How do I check a substitution solution?

Substitute both values of the ordered pair back into both original equations. If both equations are true, your solution is correct.

Related Topics

• Systems of Equations
• Systems of Equations Word Problems
• Number of Solutions in a System
• Systems of Linear Inequalities
• Multi-Step Equations