3x
- y = 12 --------(1)
2x + y= 13 --------(2)
Add equation (1) to equation (2), i.e.,
(3x - y) + (2x + y) = 12 + 13
When this is done, the terms in y cancel out and we are left
with one unknown x.
i.e., (3x - y) + (2x + y) = 12 + 13
5x = 25
x = 5
Substitute x = 5 into (1): 3(5) - y = 12
y = 3
x
= 5 and
y = 3 is the solution of the simultaneous equations.
Example 2:
Solve the simultaneous equations
13x - 6y = 20, 7x + 4y = 18
13
x - 6
y = 20 --------(1)
7x + 4y = 18
--------(2)
The coefficients of y in both equations will be numerically
equal if we multiply (1) by 2 and (2) by 3, since the LCM of 6 and 4 is 12.
(1) x 2: 26x - 12y = 40
--------(3)
(2) x 3: 21x +12y = 54
--------(4)
(3) +
(4): 47x = 94
x = 2
Substitute x = 2 into (1): 13(2) - 6y = 20
6 = 6y
y = 1
x = 2,
y =1.
