Last updated on August 27th, 2025
The substitution method is one of the methods used for solving a system of linear equations in two variables. It is useful when one of the equations is already solved, or can be rearranged for one variable in terms of the other.
The substitution method in algebra solves a system of equations having two or more variables. First, it isolates one variable in one of the equations, which forms a new expression. This expression is then substituted into the other equation. Doing so helps eliminate the other variable and solve the remaining equation with only one variable. Once solved, the value found is substituted back into the original equations to find the other variable. The answers can be verified by substituting the values of the variables in the original equations.
Aspects | Elimination method | Substitution method |
Meaning | Add or subtract the equation to cut out the one variable | Replace the one variable with the equivalent from another variable. |
When to use | When variables have the same or opposite coefficients. | When one equation is already solved or easy to solve for a variable |
Efficiency | Most efficient when equations have matching or opposite coefficients for one variable, making elimination easier. | Simple systems or when isolation is easy |
Possible risk | Might lead to tricky fractions that are harder to simplify. | Substituting increases the number of steps, making it more prone to human errors. |
Adaptability | Useful when the equations are written in different styles or formats. | Useful when equations are in standard form. |
The substitution method expresses one variable in terms of the other. To do so, the first thing done is algebraic isolation of one of the variables. Let’s say we have an equation having 2 variables x and y. The method of substitution includes algebraic isolation, substitution, solving a single variable equation, back substitution, and verification. Given below are three steps explaining this method.
Example:
Solve the equation using substitution method
x = 5 - 2y (1)
3x + y = 4 (2)
Solution
First:
Substitute equation (1) into equation (2). The equation gives x, substitute it into equation (2)
x = 5 - 2y
3x + y = 4
3(5 - 2y) + y = 4
Second, solve the equation for y
Multiply by 3 both inside the brackets
15 - 6y + y = 4
Combine the terms -6y and y we get 5y
15 - 5y = 4
Then change the coefficient
-5y = 4 - 15
-5y = - 11
y = 11/5
Third substitute y back into equation (1)
Now y = 11/5 in equation (1)
X = 5 - 2y
X = 5 - 2(11/5)
Then multiply
X = 5 - 22/5
Write 5 as a fraction with denominator 5
X = 25/5 - 22/5 = 3/5
The value of x and y is, x = 3/5, y = 11/5
We use substitution in everyday problems by swapping one unknown with an expression, so we can solve for both values more easily.
Problem-solving:
You have two choices—spending time with family or friends—and since they’re connected, you can use the substitution method to find a balance between them.
Business Planning:
When producing two related products, you can use the substitution method to figure out how many of each to make or sell to reach a target.
Planning for Trip:
While planning a trip with two parts—like driving and taking a train—if you know the speed and time of one part compared to the other, you can use substitution to figure out the total distance or time.
Time Management:
If you're dividing your time between two tasks—like studying and playing—and you know how much longer one takes than the other, you can use substitution to figure out how to split your time within the total hours you have.
Construction:
In construction, if you're using standard and special bricks and know the total number needed, you can use substitution to figure out how many of each to use, especially if standard bricks are cheaper, and you want to use more of them.
The substitution method, while helpful, can sometimes be confusing. Being aware of commonly occurring mistakes alerts students to keep them in mind and avoid them.
Solve the equation using substitution methods: x = 3y + 2, 2x + y + 12
x is 387, and y is 87
Substitute x from the first equation into the second
2(3y + 2) + y = 12
Multiply by 2 both inside the brackets
6y + 4 + y = 12
Combine the terms 6y and y we get y
7y + 4 = 12
Then change the coefficient
7y = 12 - 4
7y = 8
y = 8/7
Then find x
x = 3y + 2
x = 3(8/7) + 2
x = 24/7 + 2
Write 7 as a fraction with a denominator of 7
x = 24/7 + 14/7 = 38/7
The value of x is 38/7, and y is 8/7
Solve the equation using substitution methods: 2x − y = 3, x = y + 4
x = −1, y = −5
Substitute x:
2(y+4) − y = 3
2y + 8 − y = 3
y + 8 = 3
y = 3 − 8 = −5
Find x:
x = −5 + 4 = −1
Answer is x = −1, y = −5
Solve the equation using substitution methods: x = y + 2, 2x + y = 13
x = 5, y = 3
Substitute x into the second equation
2(y+2) + y = 13
Multiply by 2 both inside the brackets
2y + 4 + y = 13
Combining the terms 2y and y, we get y
3y + 4 = 13
Then change the coefficient and subtract 4:
3y = 9
Divide:
y = 3
Find x:
x = y + 2
x = 3 + 2 = 5
x = 5
The value of x is 5 and y is 3
Solve the equation using substitution methods: x=4y, x+y=10
x = 8, y = 2
Substitute x = 4y into the second equation:
4y + y = 10
Combine the terms 5y and y we get y
5y =10
Divide:
y = 2
Find x:
x = 4 × 2 = 8
x = 8
Solve the equation using substitution methods: x = y + 4, 2x − y = 10
x = 6, y = 2
Substitute x = y + 4:
Multiply by 2 both inside the brackets
2(y+4) − y = 10
Expand:
2y + 8 − y = 10
Then change the coefficient and subtract 8
y + 8 = 10
y = 2
Find x:
x = 2 + 4 = 6
x = 6