Solve the Linear Equation using Substitution Method

A linear equation is an equation where the highest power of the variable is always 1. Its graph is always a straight line. A linear equation in one variable has only one unknown with a degree of 1, such as: 3x + 4 = 0, 2y = 8, m + n = 5, 4a – 3b + c = 7, x/2 = 8.

There are mainly two methods for solving simultaneous linear equations:

1) Graphical method - Plotting equations and identifying their intersection.

2) Algebraic method - Further classified into:

• Substitution Method
• Elimination Method
• Cross-multiplication Method

Substitution Method

The substitution method is an algebraic technique for solving a system of linear equations with two variables.

1. Expressing one variable in terms of the other from one equation.
2. Substituting this expression into the second equation to obtain a single-variable equation.
3. Solving for the variable.
4. Substituting the solved value back into one of the original equations to find the other variable.

Steps to Solve

The following are the steps that are applied while solving a system of equations by using the Substitution Method.

Step 1: If necessary, expand the parentheses to simplify the given equation.

Step 2: Solve one of the given equations for any of the variables. Depending upon the ease of calculation, you can use any variable.

Step 3: Now, substitute the solution obtained from step 2 into the other equation.

Step 4: Now, simplify the new equation obtained by using the fundamental arithmetic operations and solve the equation for one variable.

Step 5: Finally, to find the value of the second variable, substitute the value of the variable obtained from step 4 into any of the given equations.

Now, Let us go through an example of solving a system of equations by using the substitution method, 3(x + 4) − 6y = 0 and 5x + 3y + 7 = 0.

Solution:

Step 1: By simplifying the first equation further, we get 3x − 6y + 12 = 0.

Now, the two equations are:

3x − 6y + 12 = 0  ———— (1)
5x + 3y + 7 = 0 ———— (2)

Step 2: By solving equation (1),   x = (−12 + 6y)/3 = −4 + 2y 

Step 3: Substitute the value of x obtained equation (2). i.e., we are substituting x = −4 + 2y  in the equation 5x + 3y + 7 = 0.

5(−4 + 2y) + 3y + 7 = 0

Step 4: Now, simplify the new equation obtained in the above step.

⇒ 5(−4 + 2y) + 3y + 7 = 0
⇒ −20 + 10y + 3y + 7 = 0
⇒ 13y − 13 = 0
⇒ 13y = 13
⇒ y = 13/13 ⇒ y = 1

Step 5: Now, substitute the value of y obtained in any of the given equations. Let us substitute the value of y in equation (1).

⇒ 3x − 6y + 12 = 0
⇒ 3x −6(1) + 12 = 0
⇒ 3x −6 + 12 = 0
⇒ 3x + 6 = 0
⇒ 3(x + 2) = 0
⇒ x + 2 = 0 ⇒ x = −2

Thus, by solving the given system of equations using the substitution method, we get x = −2 and y= 1.

Substitution vs Elimination Method

The substitution method and the elimination method are algebraic methods for solving simultaneous linear equations.

Related Articles

• Pair Of Linear Equations In Two Variables
• Linear Equations Formula

Solving Linear Equation using Substitution Method - Examples

Example 1: Solve: 4x−3y = 5 and 3x + y = 7, using the substitution method.

Solution: 

The given two equations are:

4x−3y = 5    ————(1)
3x + y = 7    ————(2)

Now, the solution to the given two equations can be found by the following steps:

From equation (2) we can find the value of y in terms of x, i.e.,

y = 7 − 3x

Now, substitute the value of y in equation (1). 

⇒  4x − 3(7−3x) = 5
⇒ 4x − 21+ 9x = 5
⇒ 13x = 21 + 5
⇒ 13x = 26
⇒ x = 26/13 = 2

Substitute the value of x in equation 2, 

⇒ 3(2) + y = 7
⇒ y = 7 − 6 = 1

Hence, the values of x and y are 2 and 1, respectively.

Example 2: Solve: 2m + 5n = 1 and 3m − 2n = 11 by using substitution method.

Solution: 

The given two equations are:

2m + 5n = 1  ————(1)
3m − 2n = 11   ————(2)

Now, the solution to the given two equations can be found by the following steps:

From equation (2) we can find the value of m in terms of n, i.e.,

m = (11 + 2n)/3  ————(3)

Now, substitute the value of m in equation (1). 

⇒ 2[(22 + 2n)/3] + 5n = 1
⇒ (22 + 4n)/3 + 5n = 1
⇒ [(22 + 4n) + 15n]/3 = 1
⇒ 22 + 19n = 3
⇒ 19n = 3 − 22 = −19
⇒ n = −19/19 = −1

Substitute the value of n in equation 3, 

⇒ m = (11 + 2(−1))/3
⇒ m = (11−2)/3
⇒ m = 9/3 = 3

Hence, the values of m and n are 3 and −1, respectively.

Example 3: Solve 6a − 4b = 15 and 2a + 3b = −8 by using substitution method.

Solution:

The given two equations are:

6a − 4b = 15  ————(1)
2a + 3b = −8   ————(2)

Now, the solution to the given two equations can be found by the following steps:

From equation (2) we can find the value of "a" in terms of b, i.e.,

a = (−8 − 3b)/2  ————(3)

Now, substitute the value of "a" in equation (1). 

⇒ 6[(−8 − 3b)/2) − 4b = 15
⇒ (−48 − 18b)/2 − 4b = 15
⇒ −48 − 18b − 8b = 15 × 2
⇒ −48 − 26b = 30
⇒ −26b = 30 + 48 = 78
⇒ b = −78/26 = −3

Now substitute the value of "b" in equation (3)

⇒ a = (−8 −3(−3))/2
⇒ a = (−8 + 9)/2 = 1/2
⇒ a = 0.5

Hence, the values of a and b are 0.5 and −3, respectively.

Example 4: If the sum of the two numbers is 38 and the difference between them is 12. Find the numbers using the substitution method.

Solution:

Let the two numbers be x and y.

From the given data, we can write

x + y = 38   ————(1)
x − y = 12   ————(2)

Now, the solution to the given two equations can be found by the following steps:

From equation (2) we can find the value of x in terms of y, i.e.,

x = 12 + y  ————(3)

Now, substitute the value of x in equation (1). 

⇒ 12 + y + y = 38
⇒ 12 + 2y = 38
⇒ 2y = 38 − 12 = 26
⇒ y = 26/2 = 13

 Now substitute the value of y in equation (3)

⇒ x = 12 + 13 = 25

Hence, the two given numbers are 25 and 13.

Example 5: Solve: m + n = 5 and 4m − 3n = 6 by using substitution method.

Solution:

The given two equations are:

m + n = 5  ————(1)
4m − 3n = 6   ————(2)

Now, the solution to the given two equations can be found by the following steps:

From equation (1) we can find the value of m in terms of n, i.e.,

m = 5 − n ————(3)

Now, substitute the value of m in equation (2).

⇒ 4(5−n) − 3n = 6
⇒ 20 − 4n − 3n = 6
⇒ 20 − 7n =6
⇒ 20 − 6 = 7n
⇒ 7n = 14
⇒ n = 14/7 ⇒ n = 2

Substitute the value of n in equation 1,

⇒ m + 2 = 5
⇒ m = 5 − 2 ⇒ m = 3

Hence, the values of m and n are 3 and 2, respectively.

Practice Problems - Solve the Linear Equation using Substitution Method

Question 1: Given the system of equations y = 2x + 3, x + y = 8. how would you use the substitution method to find the values of x and y?

Question 2: Solve for x and y using the substitution method: 3x - y = 2 and y = 3x - 4.

Question 3: If y = 5x - 7 and 2x + 3y = 6, use the substitution method to determine the values of x and y.

Question 4: For the equations x = 4y + 1 and 2x - 3y = 12, explain how to find x and y using the substitution method.

Question 5: How would you solve the following system using the substitution method: x - 2y = -1 and 3x + 2y = 22.