Top

Solving Linear Equations by Elimination

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

Back to Top
Elimination method uses the addition and subtraction property of equality for solving the systems of equations. Substitution method makes use of solving an equation literal for a variable. Elimination method is useful whenever one of the variables cannot solved as a simple expression of the other. Elimination method also makes the solving process easier if fractions are involved in the equations. If the coefficients of any variable are seen same in the given equations, the variable can be removed by addition or subtraction without multiplying to get the equivalent equations. Of course, you have to use the elimination method when the problem direction says so.

Solved Example

Question: Solve the system of equations  
$- 3x + 4y = 11$   
                 
$5x - 2y = 5$   

Solution:
Given
$- 3x + 4y = 11$      ..................(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

Back to Top
Elimination method is a algebraic methods used to solve a system of linear equations in two variables. Eliminate that variable whose coefficients can be made equal by easy multiplication or multiplication done on one equation only. If one equation is a multiple of the other equations, the equations are coincident. Hence the system has infinitely many solutions. On multiplying one of the equation suitably by a constant if the resulting equation varies from the second equation only by the constant term, then the two lines are parallel and the system has no solution.
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.

Step 5:
Substitute the value of the variable solved in one of the equations back and solve that equation for the other variable.

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 = 12 and y = 0.
 

Examples of Linear Systems

Back to Top
Below you could see some 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 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 x = 1 and y = -1