Substitution Method: Steps and Examples

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.

Extra practice: The Algebra 1 Unit 5 Practice Pack has printable problems, answer keys and a cheat sheet for this unit.

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

Solve one equation for one variable. Pick a variable with a coefficient of 1 or −1 if you can.
Substitute that expression into the other equation. Use parentheses.
Solve the new equation. It has only one variable.
Back-substitute. Put that value into the equation from Step 1 to find the other variable.
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 $x + y = 12$ .

y = 2x

Step 1. Look for a variable that is alone. In the first equation, $y$ is already alone.

x + 2x = 12

Step 2. Substitute $2x$ for $y$ in the second equation.

3x = 12

Step 3. Combine like terms. (See combining like terms.)

x = 4

Step 4. Divide both sides by 3.

y = 2(4) = 8

Step 5. Back-substitute $x = 4$ into $y = 2x$ .

Answer: (4, 8)

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

Example 2: Substitute an expression

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

2x + (x - 3) = 9

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

3x - 3 = 9

Step 2. Combine like terms.

3x = 12

Step 3. Add 3 to both sides.

x = 4

Step 4. Divide both sides by 3.

y = 4 - 3 = 1

Step 5. Back-substitute $x = 4$ into $y = x - 3$ .

Answer: (4, 1)

Check: $1 = 4 - 3$ and $2(4) + 1 = 8 + 1 = 9$ .

Example 3: Solve for a variable first

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

x = 7 - 2y

Step 1. In the first equation, $x$ has a coefficient of 1. Solve for $x$ by subtracting $2y$ from both sides. (See literal equations.)

3(7 - 2y) - y = 7

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

21 - 6y - y = 7

Step 3. Distribute the 3. (See distributive property.)

21 - 7y = 7

Step 4. Combine like terms.

-7y = -14

Step 5. Subtract 21 from both sides.

y = 2

Step 6. Divide both sides by $-7$ .

x = 7 - 2(2) = 7 - 4 = 3

Step 7. Back-substitute $y = 2$ into $x = 7 - 2y$ .

Answer: (3, 2)

Check: $3 + 2(2) = 3 + 4 = 7$ and $3(3) - 2 = 9 - 2 = 7$ .

Example 4: Watch the negative sign

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

4x - 3(2x - 6) = 6

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

4x - 6x + 18 = 6

Step 2. Distribute the $-3$ . A negative times a negative is positive, so $-3(-6) = 18$ .

-2x + 18 = 6

Step 3. Combine like terms.

-2x = -12

Step 4. Subtract 18 from both sides.

x = 6

Step 5. Divide both sides by $-2$ .

y = 2(6) - 6 = 12 - 6 = 6

Step 6. Back-substitute $x = 6$ into $y = 2x - 6$ .

Answer: (6, 6)

Check: $4(6) - 3(6) = 24 - 18 = 6$ and $2(6) - 6 = 12 - 6 = 6$ .

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?

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

Step 1. Let $a$ be the number of adult tickets and $c$ the number of child tickets. Write one equation for the number of tickets and one for the cost.

a = 7 - c

Step 2. Solve the first equation for $a$ . Subtract $c$ from both sides.

12(7 - c) + 8c = 68

Step 3. Substitute into the cost equation.

84 - 12c + 8c = 68

Step 4. Distribute the 12.

84 - 4c = 68

Step 5. Combine like terms.

-4c = -16

Step 6. Subtract 84 from both sides.

c = 4

Step 7. Divide both sides by $-4$ .

a = 7 - 4 = 3

Step 8. Back-substitute $c = 4$ into $a = 7 - c$ .

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

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

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

Mistake	What it looks like	How to fix it
Leaving out parentheses	$3 \cdot 7 - 2y - y = 7$	Write $3(7 - 2y)$ 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 variable	Answering “ $x = 4$ “	A solution needs both values: (4, 8).
Checking in the wrong place	Checking only in the equation you solved for y	That equation always works. Check in both original equations.

Practice Problems

Solve each system by substitution. Check your answers.

$y = 3x$ and $x + y = 16$
$y = x + 2$ and $3x + y = 18$
$x = 2y$ and $x - y = 5$
$y = -2x + 1$ and $4x + y = 9$
$x + y = 10$ and $2x - y = 2$
$2x + 3y = 13$ and $x - 2y = -4$
$3x - 2y = 4$ and $y = 2x - 1$
$y = x + 4$ and $y = x - 1$
The sum of two numbers is 25. One number is 7 more than the other. Find the numbers.
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

x + 3x = 16

Substitute $3x$ for $y$ in the second equation.

4x = 16

Combine like terms.

x = 4

Divide both sides by 4.

y = 3(4) = 12

Back-substitute into y = 3x.

Check: 12 = 3(4) and 4 + 12 = 16. Answer: (4, 12).

3x + (x + 2) = 18

Substitute x + 2 for y in the second equation.

4x + 2 = 18

Combine like terms.

4x = 16

Subtract 2 from both sides.

x = 4

Divide both sides by 4.

y = 4 + 2 = 6

Back-substitute into y = x + 2.

Check: 3(4) + 6 = 18. Answer: (4, 6).

2y - y = 5

Substitute 2y for x in the second equation.

y = 5

Combine like terms.

x = 2(5) = 10

Back-substitute into x = 2y.

Check: 10 − 5 = 5. Answer: (10, 5).

4x + (-2x + 1) = 9

Substitute −2x + 1 for y in the second equation.

2x + 1 = 9

Combine like terms.

2x = 8

Subtract 1 from both sides.

x = 4

Divide both sides by 2.

y = -2(4) + 1 = -8 + 1 = -7

Back-substitute into y = −2x + 1.

Check: 4(4) + (−7) = 16 − 7 = 9. Answer: (4, −7).

y = 10 - x

Solve the first equation for y. Subtract x from both sides.

2x - (10 - x) = 2

Substitute 10 − x for y in the second equation.

2x - 10 + x = 2

Distribute the negative sign.

3x - 10 = 2

Combine like terms.

3x = 12

Add 10 to both sides.

x = 4

Divide both sides by 3.

y = 10 - 4 = 6

Back-substitute into y = 10 − x.

Check: 4 + 6 = 10 and 2(4) − 6 = 2. Answer: (4, 6).

x = 2y - 4

Solve the second equation for x. Add 2y to both sides.

2(2y - 4) + 3y = 13

Substitute 2y − 4 for x in the first equation.

4y - 8 + 3y = 13

Distribute the 2.

7y - 8 = 13

Combine like terms.

7y = 21

Add 8 to both sides.

y = 3

Divide both sides by 7.

x = 2(3) - 4 = 6 - 4 = 2

Back-substitute into x = 2y − 4.

Check: 2(2) + 3(3) = 4 + 9 = 13 and 2 − 2(3) = −4. Answer: (2, 3).

3x - 2(2x - 1) = 4

Substitute 2x − 1 for y in the first equation.

3x - 4x + 2 = 4

Distribute the −2.

-x + 2 = 4

Combine like terms.

-x = 2

Subtract 2 from both sides.

x = -2

Divide both sides by −1.

y = 2(-2) - 1 = -4 - 1 = -5

Back-substitute into y = 2x − 1.

Check: 3(−2) − 2(−5) = −6 + 10 = 4. Answer: (−2, −5).

x + 4 = x - 1

Substitute x + 4 for y in the second equation.

4 = -1

Subtract x from both sides.

4 = −1 is false, so there is no solution. The lines are parallel.

x + y = 25

Let the numbers be x and y. Write the sum.

x = y + 7

One is 7 more than the other.

(y + 7) + y = 25

Substitute y + 7 for x in the sum.

2y + 7 = 25

Combine like terms.

2y = 18

Subtract 7 from both sides.

y = 9

Divide both sides by 2.

x = 9 + 7 = 16

Back-substitute into x = y + 7.

Check: 16 + 9 = 25 and 16 is 7 more than 9. The numbers are 16 and 9.

10.

a + p = 10

Let a be pounds of apples and p pounds of pears. Write the weight equation.

2a + 3p = 24

Write the cost equation.

a = 10 - p

Solve the weight equation for a. Subtract p from both sides.

2(10 - p) + 3p = 24

Substitute 10 − p for a in the cost equation.

20 - 2p + 3p = 24

Distribute the 2.

20 + p = 24

Combine like terms.

p = 4

Subtract 20 from both sides.

a = 10 - 4 = 6

Back-substitute into a = 10 − p.

Check: 2(6) + 3(4) = 12 + 12 = 24. Sam bought 6 pounds of apples and 4 pounds of pears.

Want more practice? The Algebra 1 Unit 5 Practice Pack has 55 pages of problems on this and the other lessons in the unit, with step-by-step answers and a cheat sheet.

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

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