Solving Linear Equations Using the Elimination Method

If an equation is written in the form ax + by + c = 0, where a, b, and c are real integers and the coefficients of x and y, i.e. a and b, are not equal to zero, it is said to be a linear equation in two variables. For example, 3x + y = 4 is a linear equation in two variables- x and y. The numbers that come before these variables are called coefficients. Thus the coefficient of x is 3 and that of y is 1.

Linear equation using Elimination Method

In the elimination method, you either add or subtract the equations to get an equation in one variable. When the coefficients of one variable are of different signs (negative in one equation, positive in the other), you add the equations to eliminate a variable and when the coefficients of one variable have the same sign (either negative in both equations or positive in both equations), you subtract the equations to eliminate a variable. It is to be noted that the variable to be eliminated needs to have the same coefficient in both equations.

The elimination method of solving a pair of linear equations is shown below, followed by the steps it entails:

Generalized example

Suppose we are given two equations of the form ax + by = c and mx + ny = q.

Given: 

ax + by = c ⇢ (1)

mx + ny = q ⇢ (2)

In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Therefore, we multiply equation (1) by m and equation (2) by a in order to equate the coefficients of x.

amx + bmy = cm

amx + any = aq

Subtracting the two equations, we obtain:

(amx – amx) + (bmy – any) = cm – aq

⇒ y(bm – an) = cm – aq

⇒ y = (cm – aq)/(bm – an)

The equation (1) can be re- written as x = c – by/a. Now substitute y = (cm – aq)/(bm – an) in the equation, x = c – by/a.

Steps to solve a linear equation using elimination method

• Make sure that the linear equations are of the form ax + by = m and cx + dy = n.
• In order to solve the given equations by elimination, the coefficients of one of the variables in both equations must be equal. Look for the numbers, which when multiplied by the coefficients of the given equations, would equate them. Just like we multiplied the above equations with m and a in order to eliminate x by making changing its coefficient in equation one from a to am and from m to am in the second equation.
• Add or subtract the equations to eliminate the variable with equal coefficients. In the above example, variable x was eliminated.
• Solve for the value of the other variable. In the example above, after eliminating x, the value of y was calculated.
• Substitute the value of the variable into any of the given equations and solve for the variable that was eliminated earlier.

Sample Problems

Question 1: Solve 2x + y = 3 and 6x − y = 9 using elimination method.

Solution:

Given: 

2x + y = 3 ⇢ (1)

6x − y = 9 ⇢ (2)

In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Therefore, multiply the first equation by -3,and multiply the second equation by 1.

−6x − 6y = −9

6x − 3y = 9

Add these equations to eliminate x:

−9y = 0

⇒ y = 0

Substitute y = 0 in equation (1):

2x + 0 = 3

⇒ x = 3/2

Thus, by elimination method, x = 3/2 and y = 0.

Question 2: Solve 4x + 2y = 10, 5x − y = 4 using elimination method.

Solution:

Given: 

4x + 2y =10 ⇢ (1)

5x − y = 4 ⇢ (2)

In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Therefore, multiply the first equation by 1,and multiply the second equation by 2.

4x + 2y = 10

10x – 2y = 8

Add these equations to eliminate y:

14x = 18

⇒ x = 18/14

⇒ x = 9/7

Substitute x = 9/7 in equation (1):

2x + 0 = 3

⇒ 4(9/7) + 2y = 10

⇒ y = 17/7

Thus, by elimination method, x = 9/7 and y = 17/7.

Question 3: Solve 9a + 2b = 6, 4a − 7b = 2 using elimination method.

Solution:

Given: 

9a + 2b = 6 ⇢ (1)

4a − 7b = 2 ⇢ (2)

In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Therefore, multiply the first equation by 7,and multiply the second equation by 2.

63a + 14b = 42

8a − 14b = 4

Add these equations to eliminate b:

71a = 46

⇒ a = 46/71

Substitute a = 46/71 in equation (1):

9(46/71) + 2b = 6

⇒ b = 6/71

Thus, by elimination method, a = 46/71 and b = 6/71.

Question 4: Solve: 3u + 2t = 8; 5u + 9t = 2 using elimination method.

Solution:

Given: 

3u + 2t = 8 ⇢ (1)

5u + 9t = 2 ⇢ (2)

In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Multiply the first equation by 5,and multiply the second equation by −3.

10t + 15u = 40

−27t − 15u = −6

Add these equations to eliminate u:

−17t = 34

⇒ t = −2

Substituting t = −2 in equation (1), we have:

3u + 2(−2) = 8

⇒ 3u = 12

⇒ u = 4

Thus, by elimination method, t = −2 and u = 4.

Question 5: Solve 7p + 4q = 7, 8p + 5q = 5 using elimination method.

Solution:

Given: 

3u + 2t = 8 ⇢ (1)

5u + 9t = 2 ⇢ (2)

In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Multiply the first equation by 5, and multiply the second equation by −4.

35p + 20q = 35

−32p − 20q = −20

Add these equations to eliminate q:

3p = 15

⇒ p = 15/3

⇒ p = 5

Substituting p = 5 in equation (1), we have:

7(5) + 4q = 7

⇒ 4q = -28

⇒ q = -7

Thus, by elimination method, p = 5 and q = -7.