Solve the following simultaneous equations: a + 3b = 12, 4a - 3b = 18 (KS3 2010 Paper 2)

So with simultaneous equations we want to try and eliminate one of the unknowns we have. In this case, they are a and b. If we label the two equations 1 and 2 we have to find a way to reduce the equations to find one of the letters. Looking at the equations we can see that one has a minus 3b and the other has a plus 3b so if we add them together that makes 0. So adding the equations and matching up the a terms, the b terms and the number term at the end leaves us with 5a = 30. From here we can see that dividing both sides of this equation by 5 gives a = 6. Subbing a = 6 back into the first equation gives 6 + 3b = 12. Subtracting both sides of this equation by 6 leaves 3b = 6. Dividing both sides by 3 leaves b = 2 and so the answer is a = 6 and b = 2