Elimination Method
In Elimination Method, one variable either x or y is eliminated to
find the value of other variable. To solve the pair of simultaneous equations
by this method, we need to make coefficients of any one variable equal by
multiplying or dividing the equations by a least number so that the coefficient
of x or y become equal in two new equations.
Then by adding or
subtracting according to these equal coefficients have opposite or same sign
respectively, we get an equation which will have only one variable.
Simplification will give the value of that variable.
After solving the value of
either x or y obtained from above steps is substituted in any one of the given
equations to get the value of the other variable.
Steps:
i. Convert one
variable of both equations with the same coefficient.
ii. Eliminate the
variable having the same coefficient by subtracting or adding.
iii. Solve the
equation having one variable.
iv. Substitute the
value of one variable so obtained in any one of the given equations to
calculate the value of other variable.
This process of Elimination Method will be clear by the
following worked out examples.
Workout Examples
Example 1: Solve: x + 2y = 5 and 5x – 2y
= 13 by elimination method.
Solution: Here,
x + 2y = 5 …………. (i)
5x – 2y = 13 ………….. (ii)
Adding equations (i) and (ii) we
get,
or, x
= 18/6
or, x
= 3
Now, putting x = 3 in equation (i)
we get,
3 +2y = 5
or, 2y
= 5 – 3
or, 2y
= 2
or, y
= 2/2
or, y
= 1
Hence, x = 3 and y = 1.
Example 2: Solve: 3x + 5y = 7 and 7x +
10y = 13 by elimination method.
Solution: Here,
3x + 5y = 7 …………. (i) ×
2
7x + 10y = 13 …………..
(ii)
Multiplying
equation (i) by 2 and subtracting from equation (ii) we get,
Now,
putting x = –1 in equation (i) we get,
3
× –1 + 5y = 7
or, –3 + 5y = 7
or, 5y = 7 + 3
or, 5y = 10
or, y = 10/5
or, y = 2
Hence, x = –1 and y = 2.
Example 3: Solve: 5x + 3y = 5 and 3x – 4y
= 32 by elimination method.
Solution: Here,
5x + 3y = 5 …………. (i) × 4
3x – 4y = 32 ………….. (ii)
× 3
Multiplying equation (i) by 4 and (ii) by 3 and then adding
we get,
or, x = 4
Now,
putting x = 4 in equation (i) we get,
5
× 4 + 3y = 5
or, 20 + 3y = 5
or, 3y = 5 – 20
or, 3y = –15
or, y = –15/3
or, y = –5
Hence, x = 4 and y = –5.
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questions or problems regarding the solving of simultaneous linear equations by
elimination method here.
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