Substitution Method for Systems of Equations

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The substitution method solves a system of equations by replacing one variable with an equal expression. Solve one equation for x or y, then substitute that expression into the other equation and solve. Put that answer back into either equation to find the other variable, and check both.

Before this lesson: You should know what a system of equations is, how to use the distributive property and how to solve multi-step equations.

How the Substitution Method Works

Look at this system:

\begin{cases}
y = 2x \\
x + y = 12
\end{cases}

The first equation says y is the same as 2x. So in the second equation, you can swap y for 2x. That leaves an equation with only x in it, which you already know how to solve.

That swap is the whole idea. “Substitute” means “put in place of.”

Steps for the Substitution Method

  1. Solve one equation for one variable. Pick a variable with a coefficient of 1 or −1 if you can.
  2. Substitute that expression into the other equation. Use parentheses.
  3. Solve the new equation. It has only one variable.
  4. Back-substitute. Put that value into the equation from Step 1 to find the other variable.
  5. Check the pair in both original equations.

Substitution is a good choice when one variable is already by itself, like y = 2x - 6, or is easy to get by itself.

Worked Examples

Example 1: One variable is already alone

Solve the system y = 2x and x + y = 12.

Step 1. y is already alone: y = 2x.

Step 2. Substitute 2x for y in the second equation, then solve.

\begin{aligned}
x + 2x &= 12 \\
3x &= 12 \\
x &= 4
\end{aligned}

Step 3. Back-substitute: y = 2(4) = 8.

Check: 8 = 2(4) and 4 + 8 = 12. Both are true.

Answer: (4, 8)

Example 2: Substitute an expression

Solve the system y = x - 3 and 2x + y = 9.

Step 1. Substitute x - 3 for y in the second equation.

\begin{aligned}
2x + (x - 3) &= 9 \\
3x - 3 &= 9 \\
3x &= 12 \\
x &= 4
\end{aligned}

Step 2. Back-substitute: y = 4 - 3 = 1.

Check: 1 = 4 - 3 and 2(4) + 1 = 9.

Answer: (4, 1)

Example 3: Solve for a variable first

Solve the system x + 2y = 7 and 3x - y = 7.

Step 1. In the first equation, x has a coefficient of 1. Solve for x: x = 7 - 2y.

Step 2. Substitute 7 - 2y for x in the second equation. Distribute the 3.

\begin{aligned}
3(7 - 2y) - y &= 7 \\
21 - 6y - y &= 7 \\
21 - 7y &= 7 \\
-7y &= -14 \\
y &= 2
\end{aligned}

Step 3. Back-substitute: x = 7 - 2(2) = 3.

Check: 3 + 2(2) = 7 and 3(3) - 2 = 7.

Answer: (3, 2)

Example 4: Watch the negative sign

Solve the system 4x - 3y = 6 and y = 2x - 6.

Step 1. Substitute 2x - 6 for y. Keep the parentheses, because the −3 multiplies both terms.

\begin{aligned}
4x - 3(2x - 6) &= 6 \\
4x - 6x + 18 &= 6 \\
-2x + 18 &= 6 \\
-2x &= -12 \\
x &= 6
\end{aligned}

Step 2. Back-substitute: y = 2(6) - 6 = 6.

Check: 4(6) - 3(6) = 24 - 18 = 6 and 2(6) - 6 = 6.

Answer: (6, 6)

Example 5: Word problem

Adult movie tickets cost $12, and child tickets cost $8. A family buys 7 tickets for $68. How many of each did they buy?

Step 1. Let a be the number of adult tickets and c the number of child tickets.

\begin{cases}
a + c = 7 & \text{number of tickets} \\
12a + 8c = 68 & \text{total cost}
\end{cases}

Step 2. Solve the first equation for a: a = 7 - c.

Step 3. Substitute into the cost equation and solve.

\begin{aligned}
12(7 - c) + 8c &= 68 \\
84 - 12c + 8c &= 68 \\
84 - 4c &= 68 \\
-4c &= -16 \\
c &= 4
\end{aligned}

Step 4. Back-substitute: a = 7 - 4 = 3.

Check: 3 + 4 = 7 tickets and 12(3) + 8(4) = 36 + 32 = 68 dollars.

Answer: They bought 3 adult tickets and 4 child tickets.

When the Variables Disappear

Sometimes both variables cancel out when you substitute. What’s left tells you about the system.

  • A false statement, like 4 = -1, means the system has no solution. The lines are parallel.
  • A true statement, like 0 = 0, means the system has infinitely many solutions. Both equations describe the same line.

You’ll practice these cases in Number of Solutions.

Common Mistakes

MistakeWhat it looks likeHow to fix it
Leaving out parentheses3 \cdot 7 - 2y - y = 7Write 3(7 - 2y) and distribute to both terms.
Losing a negative-3(2x - 6) = -6x - 18-3 \cdot (-6) = +18, so it is -6x + 18.
Stopping after one variableAnswering “x = 4“A solution needs both values: (4, 8).
Checking in the wrong placeChecking only in the equation you solved for yThat equation always works. Check in both original equations.

Practice Problems

Solve each system by substitution. Check your answers.

  1. y = 3x and x + y = 16
  2. y = x + 2 and 3x + y = 18
  3. x = 2y and x - y = 5
  4. y = -2x + 1 and 4x + y = 9
  5. x + y = 10 and 2x - y = 2
  6. 2x + 3y = 13 and x - 2y = -4
  7. 3x - 2y = 4 and y = 2x - 1
  8. y = x + 4 and y = x - 1
  9. The sum of two numbers is 25. One number is 7 more than the other. Find the numbers.
  10. A farm stand sells apples for $2 per pound and pears for $3 per pound. Sam buys 10 pounds of fruit for $24. How many pounds of each did Sam buy?
Show answers
  1. x + 3x = 16, so x = 4 and y = 12. Answer: (4, 12).
  2. 3x + (x + 2) = 18, so 4x = 16 and x = 4. Then y = 6. Answer: (4, 6).
  3. 2y - y = 5, so y = 5 and x = 10. Answer: (10, 5).
  4. 4x + (-2x + 1) = 9, so 2x = 8 and x = 4. Then y = -2(4) + 1 = -7. Answer: (4, −7).
  5. y = 10 - x. Then 2x - (10 - x) = 2, so 3x = 12 and x = 4. Then y = 6. Answer: (4, 6).
  6. x = 2y - 4. Then 2(2y - 4) + 3y = 13, so 7y - 8 = 13 and y = 3. Then x = 2. Answer: (2, 3).
  7. 3x - 2(2x - 1) = 4, so -x + 2 = 4 and x = -2. Then y = 2(-2) - 1 = -5. Answer: (−2, −5).
  8. x + 4 = x - 1 gives 4 = -1, which is false. There is no solution.
  9. x + y = 25 and x = y + 7. Then 2y + 7 = 25, so y = 9 and x = 16. The numbers are 16 and 9.
  10. a + p = 10 and 2a + 3p = 24. Then 2(10 - p) + 3p = 24, so 20 + p = 24 and p = 4. Sam bought 6 pounds of apples and 4 pounds of pears.

Substitution Method FAQ

What is the substitution method?

The substitution method solves a system by getting one variable alone in one equation and putting that expression into the other equation. This leaves one equation with one variable.

When should you use substitution instead of elimination?

Use substitution when a variable is already alone or has a coefficient of 1 or −1. If every variable has a larger coefficient, the Elimination Method is usually faster.

What does it mean when you get 0 = 0?

A true statement like 0 = 0 means both equations describe the same line. The system has infinitely many solutions.

Summary

  • Solve one equation for a variable, then substitute that expression into the other equation.
  • Use parentheses when you substitute, and distribute carefully.
  • Back-substitute to find the second variable.
  • Check your answer in both original equations.
  • A false statement means no solution. A true statement like 0 = 0 means infinitely many solutions.

Previous lesson: Systems of Equations

Next lesson: Elimination Method

Back to the Algebra 1 Index

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