Substitution and elimination are the common algebraic methods used to solve a system of linear equations in two variables. Against the graphical method of solving a system which often can gives only an approximate solution, the algebraic methods yield exact solution of the system. In elimination method, we add the both equations together so that one of
the variables eliminated. Sometimes it is necessary to multiply one or
both equations by a constant in order for the terms to have opposite
signs. So when we add the equations, one of the variables will be
eliminated.

**Elimination method**involve solving for one of the variable by eliminating the other variable. This method also named as addition method. Multiply the original equations by a constant before it eliminate one of the variables by adding or subtracting the equations.## Solving Linear Equations by Elimination

Solved Example

**Question:**Solve the system of equations

$- 3x + 4y = 11$

$5x - 2y = 5$

**Solution:**

Given

$- 3x + 4y = 11$ ..................(1)

$5x - 2y = 5$ ...................(2)

We can eliminate any one of the variables x or y. But for the above system it is easier to

To eliminate y, multiply each side of equation (2) by 2

$2(5x - 2y) = 2(5)$

$10x - 4y = 10$ ........................(3)

(Distributive Property)

Add equation (1) and equation (3)

- 3x + 4y + 10x - 4y = 11 + 10

=> -3x + 10x = 21

=> 7x = 21

=> $x = 3$

Put x = 3 in equation (2)

=> 5 * 3 - 2y = 5

=> 15 - 2y = 5

=> -2y = 5 - 15 = -10

=> y = 5

Hence the solution to the system is

$- 3x + 4y = 11$ ..................(1)

$5x - 2y = 5$ ...................(2)

**Step 1:**We can eliminate any one of the variables x or y. But for the above system it is easier to

**eliminate y**. By multiplying the second equation by 2 we can make coefficient of y as -4.To eliminate y, multiply each side of equation (2) by 2

$2(5x - 2y) = 2(5)$

$10x - 4y = 10$ ........................(3)

(Distributive Property)

**Step 2:**Add equation (1) and equation (3)

- 3x + 4y + 10x - 4y = 11 + 10

=> -3x + 10x = 21

=> 7x = 21

=> $x = 3$

**Step 3:**Put x = 3 in equation (2)

=> 5 * 3 - 2y = 5

=> 15 - 2y = 5

=> -2y = 5 - 15 = -10

=> y = 5

Hence the solution to the system is

**x = 3 and y = 5**.## Solve Using the Elimination Method

**Steps for Solving Linear Equation by Elimination:**

**Step 1:**Write the equations in standard form, that is ax+by = c.

**Step 2:**Multiply or divide one or both the equations so that the coefficients of one of the variables is equal in both the equations.

**Step 3:**Add the two equations to get rid of the variable if the coefficients are of equal and opposite. Subtract one equation from the other, if the coefficients are of same sign.

**Step 4:**Solve the resulting equation in one variable.

**Substitute the value of the variable solved in one of the equations back and solve that equation for the other variable.**

Step 5:

Step 5:

Solved Example

**Question:**Solve the system of equations

$x + 2y = 12$

$2x + 6y = 24$

**Solution:**

Given

x + 2y = 12 ....................(1)

2x + 6y = 24 .....................(2)

Step 1:

Eliminate any one of the variables x or y.

To eliminate y, multiply each side of equation (1) by 3

=> 3(x + 2y) = 12 * 3

=> 3x + 6y = 36 ................(3)

Step 2:

Subtract equation (2) from equation (3) to eliminate y.

=> 3x + 6y - (2x + 6y) = 36 - 24

=> 3x + 6y - 2x - 6y = 12

=> x = 12

Step 3:

Put x = 12 in equation (1)

=> 12 + 2y = 12

=> 2y = 12 - 12 = 0

=> y = 0

Hence the solution to the system is

x + 2y = 12 ....................(1)

2x + 6y = 24 .....................(2)

Step 1:

Eliminate any one of the variables x or y.

To eliminate y, multiply each side of equation (1) by 3

=> 3(x + 2y) = 12 * 3

=> 3x + 6y = 36 ................(3)

Step 2:

Subtract equation (2) from equation (3) to eliminate y.

=> 3x + 6y - (2x + 6y) = 36 - 24

=> 3x + 6y - 2x - 6y = 12

=> x = 12

Step 3:

Put x = 12 in equation (1)

=> 12 + 2y = 12

=> 2y = 12 - 12 = 0

=> y = 0

Hence the solution to the system is

**x = 12 and y = 0**.## Examples of Linear Systems

## Solved Examples

**Question 1:**Solving a system of linear equations by addition method

2x - 5y = -2

3x - 2y = 5

**Solution:**

Given

2x - 5y = -2 ..........................(1)

3x - 2y = 5 ..........................(2)

Step 1:

Eliminate any one of the variables x or y.

To eliminate x, multiply each side of equation (1) by 3 and equation (2) by 2

(1) => 3(2x - 5y) = -2 * 3

=> 6x - 15y = - 6 .....................(3)

(2) => 2(3x - 2y) = 5 * 2

=> 6x - 4y = 10 ......................(4)

Step 2:

Coefficient of x is same in both the equations. To eliminate x, subtract (3) from (4)

=> 6x - 4y - (6x - 15y) = 10 - (-6)

=> 6x - 4y - 6x + 15y = 10 + 6

=> 11y = 16

=> y = $\frac{16}{11}$

Step 3:

Put y = $\frac{16}{11}$ in equation (1)

=> 2x - 5 * $\frac{16}{11}$ = -2

=> 2x - $\frac{80}{11}$ = -2

=> 2x = -2 + $\frac{80}{11}$

=> 2x = $\frac{58}{11}$

=> x = $\frac{29}{11}$

Hence the solution to the system is

2x - 5y = -2 ..........................(1)

3x - 2y = 5 ..........................(2)

Step 1:

Eliminate any one of the variables x or y.

To eliminate x, multiply each side of equation (1) by 3 and equation (2) by 2

(1) => 3(2x - 5y) = -2 * 3

=> 6x - 15y = - 6 .....................(3)

(2) => 2(3x - 2y) = 5 * 2

=> 6x - 4y = 10 ......................(4)

Step 2:

Coefficient of x is same in both the equations. To eliminate x, subtract (3) from (4)

=> 6x - 4y - (6x - 15y) = 10 - (-6)

=> 6x - 4y - 6x + 15y = 10 + 6

=> 11y = 16

=> y = $\frac{16}{11}$

Step 3:

Put y = $\frac{16}{11}$ in equation (1)

=> 2x - 5 * $\frac{16}{11}$ = -2

=> 2x - $\frac{80}{11}$ = -2

=> 2x = -2 + $\frac{80}{11}$

=> 2x = $\frac{58}{11}$

=> x = $\frac{29}{11}$

Hence the solution to the system is

**x =**$\frac{29}{11}$**and y =**$\frac{16}{11}$.**Question 2:**Solving a system of linear equations by elimination method

5x + 3y = 2

x - 2y = 3

**Solution:**

Given

5x + 3y = 2 ..................(1)

x - 2y = 3 ...................(2)

Step 1:

Eliminate any one of the variables x or y.

To eliminate x, multiply each side of equation (2) by 5

(2) => 5(x - 2y) = 3 * 5

=> 5x - 10y = 15 ..................(3)

Step 2:

Coefficient of x is same in both the equations. To eliminate x, subtract (3) from (1)

=> 5x + 3y - (5x - 10y) = 2 - 15

=> 5x + 3y - 5x + 10y = -13

=> 13y = -13

=> y = -1

Step 2:

Put y = -1 in equation (1)

=> 5x + 3 * -1 = 2

=> 5x - 3 = 2

=> 5x = 5

=> x = 1

Hence the solution to the system is

5x + 3y = 2 ..................(1)

x - 2y = 3 ...................(2)

Step 1:

Eliminate any one of the variables x or y.

To eliminate x, multiply each side of equation (2) by 5

(2) => 5(x - 2y) = 3 * 5

=> 5x - 10y = 15 ..................(3)

Step 2:

Coefficient of x is same in both the equations. To eliminate x, subtract (3) from (1)

=> 5x + 3y - (5x - 10y) = 2 - 15

=> 5x + 3y - 5x + 10y = -13

=> 13y = -13

=> y = -1

Step 2:

Put y = -1 in equation (1)

=> 5x + 3 * -1 = 2

=> 5x - 3 = 2

=> 5x = 5

=> x = 1

Hence the solution to the system is

**x = 1 and y = -1**.