In previous activities, we have used the 'elimination' method to solve two simultaneous equations. However, in some situations, another method may be preferred. This is the 'substitution' method in which one equation is substituted into the other. The following example will demonstrate this method.
Solve the equations
2x + 3y = 30 (1)
y = 3x - 1 (2)
Notice that equation (2) is not written in the same form as equation (1), it is written with y as the subject. This makes it more suitable for the substitution method. So, we substitute equation (2) into equation (1). This means we replace the 'y' in (1) with '3x - 1' as follows:
2x + 3(3x - 1) = 30.
Now, this equation only has one variable, x, so it can be solved. First, expand the brackets:
2x + 9x - 3 = 30
11x - 3 = 30
11x = 33
x = 3
Now, substitute x = 3 into either (1) or (2) whichever you think is easiest. Here, we substitute into (2)
y = 3 x 3 - 1
y = 8
As before we can check our solutions using equation (1)
2 x 3 + 3 x 8 = 30
6 + 24 = 30 [correct]
So x = 3 and y = 8.
The substitution method is to be preferred over the elimination method when one equation is written in the form y = ..... or x = ...... and the other is in the form ax + by = c where a, b, c are constants.
Now its time for you to try some.