# Method of Elimination

Earlier we have seen about linear equation in two variables and an overview of solution of linear equation in two variables. From this topic onwards, we will deal mostly with solution of linear equation in two variables in detail. Under this topic we will get to learn about one of the methods of solving linear equations in two variables, i.e., method of elimination of one variable. In this method of elimination of one variable we will be eliminating of the variables out of two variables so as to find out the value of one variable and then substituting it into one of the equations to get the value of another variable.

Following are the steps involved in method of elimination:

Step I: Look carefully for the two equations as it is requirement of minimum of two equations for solving linear equations in two variables.

Step II: Have a close look at each coefficient of the variables and check for the least possible common multiple of the any of the coefficients.

Step III: Multiply both equations on both sides with constants such that one of the coefficients of any one variable becomes same and eliminate this variable having same coefficient by either adding or subtracting both equations.

Step IV: While we add or subtract these equations, one of the variable gets eliminated and we are left with linear equation in one variable. We can solve this linear equation in one variable.

Step V: As we get value of any one variable we put it into any of the given equations to solve for other variable.

This method is known as method of elimination as one of the variable is getting eliminated in order for solving the equation.

To understand the concept in a better way, let us have a look at the following examples.

1. Solve for the unknown variables ‘x’ and ‘y’:

x + 2y = 15 ……….…….. (i)

2x + y = 20……….…….. (ii)

Solution:

Solving these equations by method of elimination:

Looking at the coefficients of the variables of the given equations, we find that we can multiply equation (i) by 2 on both the sides so as to make coefficients of ‘x’ equal.

2x + 4y = 30 .......... (iii)

Subtracting equation (ii) from equation (iii), we get

3y = 30 – 20

⟹ 3y = 10

⟹ y = 10/3

Substituting y = 10/3 in equation (ii), we get

2x + 10/3 = 20

⟹ 2x = 20 – 10/3

⟹ 2x = 50/3

⟹ x = 50/6

⟹ x = 25/3

Hence, x = 25/3 and y = 10/3.



2. Solve for ‘x’ and ‘y’:

2x + 3y = 12 .......... (i)

3x + 2y = 15 .......... (ii)

Solution:

If we look closely at the coefficients of the variables of both the equations, we will find that we can either multiply equation (i) by 3 on both the sides and equation (ii) by 2 on both the sides to make coefficient of ‘x’ equal and solve the equations by eliminating ‘x’ from both the equations or we can multiply equation (i) by 2 and equation (ii) by 3 on both the sides to make the coefficient of ‘y’ equal and then solve the equations by eliminating ‘y’ from both the equations.

Let us use first method.

Multiply equation (i) by 3 on both the sides and equation (ii) by 2 on both the sides to make coefficient of ‘x’ equal and solve the equations by eliminating ‘x’ from both the equations, we get

6x + 9y = 36 .......... (iii)

6x + 4y = 30 .......... (iv)

Subtract equation (iv) from equation (iii), we get

9y – 4y = 36 – 30

⟹ 5y = 6

⟹ y = 6/5

Substituting y = 6/5 in equation (i), we get

2x + 18/5 =12

⟹ 2x = 12 – 18/5

⟹ 2x = 42/5

⟹ x = 42/10

⟹ x = 21/5

Hence, x = 21/5 and y = 6/5.

3. Solve for ‘x’ and ‘y’:

5x + y = 30 .......... (i)

x +  y = 10 .......... (ii)

Solution:

As we can see that coefficients of ‘y’ in both the equations are equal. So, we don’t need to multiply any one the equations with any constant. We can directly go for elimination of ‘y’ from both the equations and solve the given equations.

On subtracting equation (ii) from equation (i), we get

5x – x = 30 - 10

⟹ 4x = 20

⟹ x = 20/4

⟹ x = 5.

Substituting x = 5 in equation (ii), we get,

5 + y = 10

⟹ y = 10 – 5

⟹ y = 5.

Hence, x = 5 and y = 5.

4. Solve for ‘x’ and ‘y’:

5x + 2y = 11 ............. (i)

2x +  y = 13 ............ (ii)

Solution:

Looking at the given linear equations we find that we can either multiply eq. (i) by 2 and equation (ii) by 5 on both sides to make the coefficients of ‘x’ equal and then solve it by eliminating ‘x’ from both the equations or we can multiply eq. (ii) by 2 on both the sides and eliminate ‘y’ and solve the equations further to get the values of ‘x’ and ‘y’.

Going by second method as it is easy and short to eliminate ‘y’ from both the equations.

Multiplying eq. (ii) by 2 on both the sides, we get

4x + 2y = 26 ............. (iii)

Subtracting eq. (i) from eq. (iii), we get;

4x – 5x = 26 – 11

⟹ - x = 15

⟹ x = -15

substituting x = -15 in eq. (ii), we get;

2(-15) + y = 13

⟹ -30 + y =13

⟹ y = 13 + 30

⟹ y = 43.

Hence, x = -15 and y = 43.

Similarly, other problems could be solved using method of elimination.