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**

Now, simplify

**11x - 3 = 30**

Now, solve

**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.