# Comparison Method for Solving System of Equations | How to Solve Simultaneous Equations Using Comparison?

Students can solve the system of linear equations by using the comparison method. Here, you need to compare a pair of equations and find the value of the variables. You can find a detailed explanation of solving the simultaneous linear equations in the following sections. The comparison method is one of the easiest ways to solve a pair of equations. So, follow the below guidelines and solve the questions.

## How to Solve Simultaneous Equations Using Comparison Method?

Simultaneous Equations are nothing but two or more equations having two or more quantities. There are many ways to solve the system of linear equations, but we are using the comparison method to get the answer easily. Check the steps and get the solution.

• Let us take (i) and (ii) as the equations numbers.
• Express both equations x in terms of y.
• And consider obtained equations as (iii), (iv).
• Equate the obtained equations i.e the values of x in the equation forming the equation in y.
• Solve the linear equation in y and simplify it.
• Substitute the obtained y value in any one equation and solve it.
• Thus, you will get the solution using the comparison method.

### Solved Examples on System of Linear Equations

Example 1.

Solve the pair of simultaneous equations 5x + 3y = 9, 2x – 3y = 12 using the comparison method with detailed explanation.

Solution:

Given the system of linear equations are

5x + 3y = 9 —— (i)

2x – 3y = 12 ——- (ii)

Express of x in terms of y.

From Equation (i) 5x + 3y = 9, we get

Subtract 3y from both sides

5x + 3y – 3y = 9 – 3y

5x = 9 – 3y

Divide both sides of the equation by 5.

5x/5 = (9 – 3y)/5

x = (9 – 3y)/5

Therefore, x = (9 – 3y)/5 —— (iii)

From equation (ii) 2x – 3y = 12, we get

Add 3y to both sides of the equation.

2x – 3y + 3y = 12 + 3y

2x = 12 + 3y

Divide both sides by 2.

2x/2 = (12 + 3y)/2

x = (12 + 3y)/2

Therefore, x = (12 + 3y)/2 —– (iv)

Equate the values of x in equation (iii) and equation (iv) forming the equation in y

From equations (iii) and (iv), we get

(9 – 3y)/5 = (12 + 3y)/2

Solve the above linear equation

Cross multiply the fractions.

2(9 – 3y) = 5(12 + 3y)

18 – 6y = 60 + 15y

18 – 60 = 15y + 6y

-42 = 21y

y = -42/21

y = -2

Substituting the value of y in equation (iii) or equation (iv), find the value of x

Put y = -2 in equation (iii), we get

x = (9 – 3(-2))/5

x = (9 + 6)/5

= 15/5

x = 3

Required solution of the two equations

Therefore, x = 3, y = -2.

Therefore, we have compared the values of x obtained from equation (i) and (ii) and formed an equation in y, so this method of solving simultaneous equations is known as the comparison method. Similarly, comparing the two values of y, we can form an equation in x.

Example 2.

Solve the system of linear equations 5x + y = 10, 7x – 3y + 8 = 0 using comparison method.

Solution:

Given pair of system of linear equations are

5x + y = 10 —- (i)

7x – 3y + 8 = 0 —– (ii)

Express x in terms of y

From equation (i) 5x + y = 10, we get

5x = 10 – y

x = (10 – y)/5

From equation (ii) 7x – 3y + 8 = 0, we get

7x – 3y = -8

7x = 3y – 8

x = (3y – 8)/7

Equate the values of x obtained from both equations.

(10 – y)/5 = (3y – 8)/7

Solve the equation in y.

7(10 – y) = 5(3y – 8)

70 – 7y = 15y – 40

70 + 40 = 15y + 7y

110 = 22y

y = 110/22

y = 5

Putting the value of y in equation (i)

Substitute y = 5 in equation (i), we get

5x + 5 = 10

5x = 10 – 5

5x = 5

x = 5/5

x = 1

Required solution of the two equations

Therefore, x = 1, y = 5.

Example 3.

Solve the linear equations 2x + y = 10 and x + y = 4 by comparison method.

Solution:

Given simultaneous linear equations are

2x + y = 10 —- (i)

x + y = 4 —- (ii)

express y in terms of x.

From equation (i), we can write

y = 10 – 2x

From equation (ii), we can write

y = 4 – x

Equate both the values of y obtained from equations (i) and (ii)

10 – 2x = 4 – x

10 – 4 = 2x – x

x = 6

Substitute x = 6 in the equation (ii)

6 + y = 4

y = 4 – 6

y = -2

Required solution of the two equations

Therefore, x = 6, y = -2.

Example 4.

Find the solution to the following simultaneous equations.

4x + 3y = 14 and 5x + 7y = 11.

Solution:

Given linear equations are

4x + 3y = 14 —– (i)

5x + 7y = 11 —— (ii)

Express x in terms of y for both equations.

From equation (i) 4x + 3y = 14, we can write

4x = 14 – 3y

x = (14 – 3y)/4

From equation (ii) 5x + 7y = 11, we can write

5x = 11 – 7y

x = (11 – 7y)/5

Equate the values of x obtained from equations (i), (ii)

(14 – 3y)/4 = (11 – 7y)/5

5(14 – 3y) = 4(11 – 7y)

70 – 15y = 44 – 28y

28y – 15y = 44 – 70

13y = -26

y = -26/13

y = -2

Substitute y = -2 in the equation (i) and simplify it to get the x value.

Put y = -2 in 4x + 3y = 14

4x + 3(-2) = 14

4x – 6 = 14

4x = 14 + 6

4x = 20

x = 20/4

x = 5

Therefore the required solutions are x = 5, y = -2.

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