Elimination Method



Follow the steps to solve the system of linear equations by using the elimination method:



(i) Multiply the given equation by suitable constant so as to make the coefficients of the variable to be eliminated equal.

(ii) Add the new equations obtained if the terms having the same coefficient are opposite signs and subtract if they are of the same sign.

(iii) Solve the equation thus obtained.

(iv) Substitute the value found in any one the given equations.

(v) Solve it to get the value of the other variable.


Worked-out examples on elimination method:

1. Solve the system of equation 2x + y = -4 and 5x – 3y = 1 by the method of elimination.

Solution:

The given equations are:
2x + y = -4       …………… (i)
5x – 3y = 1       …………… (ii)

Multiply equation (i) by 3, we get;
{2x + y = -4} …………… {× 3}
6x + 3y = -12       …………… (iii)

Adding (ii) and (iii), we get;

Elimination method



or, x = -11/11
or, x = -1

Substituting the value of x = -1 in equation (i), we get;
2 × (-1) + y = -4
-2 + y = -4
-2 + 2 + y = -4 + 2
y = -4 + 2
y = -2
Therefore, x = -1 and y = -2 is the solution of the system of equations 2x + y = -4 and 5x – 3y = 1



2. Solve the system of equation 2x + 3y = 11, x + 2y = 7 by the method of elimination.

Solution:

The given equations are:
2x + 3y = 11       …………… (i)
x + 2y = 7       …………… (ii)

Multiply the equation (ii) by 2, we get
{x + 2y = 7}       …………… (× 2)
2x + 4y = 14       …………… (iii)

Subtract equation (i) and (ii), we get

method of elimination



Substituting the value of y = 3 in equation (i), we get
2x + 3y = 11
or, 2x + 3 × 3 = 11
or, 2x + 9 = 11
or, 2x + 9 – 9 = 11 – 9
or, 2x = 11 – 9
or, 2x = 2
or, x = 2/2
or, x = 1
Therefore, x = 1 and y = 3 is the solution of the system of the given equations.


3. Solve 2a – 3/b = 12 and 5a – 7/b = 1

Solution:

The given equations are:
2a – 3/b = 12       …………… (i)
5a – 7/b = 1       …………… (ii)

Put 1/b = c, we have
2a – 3c = 12       …………… (iii)
5a – 7c = 1       …………… (iv)

Multiply equation (iii) by 5 and (iv) by 2, we get
10a – 15c = 60       …………… (v)
10a + 14c = 2       …………… (vi)

Subtracting (v) and (vi), we get

Subtracting



or, c = 58 /-29
or, c = -2
But 1/b = c
Therefore, 1/b = -2 or b = -1/2

Subtracting the value of c in equation (v), we get
10a – 15 × (-2) = 60
or, 10a + 30 = 60
or, 10a + 30 - 30= 60 - 30
or, 10a = 60 – 30
or, a = 30/10
or, a = 3
Therefore, a = 3 and b = 1/2 is the solution of the given system of equations.


4. x/2 + 2/3 y = -1 and x – 1/3 y = 3

Solution:

The given equations are:
x/2 + 2/3 y = -1       …………… (i)
x – 1/3 y = 3       …………… (ii)

Multiply equation (i) by 6 and (ii) by 3, we get;
3x + 4y = -6       …………… (iii)
3x – y = 9       …………… (iv)
Solving (iii) and (iv), we get;

elimination



or, y = -15/5
or, y = -3

Subtracting the value of y in (ii), we get;
x – 1/3 × -3 = 3
or, x + 1 = 3
or, x = 3 – 1
or, x = 2
Therefore, x = 2 and y = -3 is the solution of the equation.
x/2 + 2/3 y = -1 and x - y/3 = 3



Simultaneous Linear Equations

  • Simultaneous Linear Equations
  • Comparison Method
  • Elimination Method
  • Substitution Method
  • Cross-Multiplication Method
  • Solvability of Linear Simultaneous Equations
  • Pairs of Equations
  • Word Problems on Simultaneous Linear Equations
  • Practice Test on Word Problems Involving Simultaneous
    Linear Equations


  • Simultaneous Linear Equations - Worksheets
  • Worksheet on Simultaneous Linear Equations
  • Worksheet on Problems on Simultaneous Linear
    Equations


  • 8th Grade Math Practice

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