a) This question tests you on using fractions, brackets and quadratic equations. We should start by trying to make it into a form we know how to deal with. This will be by the balance method. If we multiply both sides through by (x-2) we can cancel the bottom portion of the first fraction.

This will give 6-(x-2)(2/x+3)=(x-2)

b)The next part is similar. We should now multiply both sides by (x+3). This will give 6(x+3)-2(x-2)=(x-2)(x+3).

c)Now if we expand the brackets on the left side we will get: 6x+18-2x+4 (be careful of the minus signs!)

Tidying this up will give: 4x+22

d)Now expanding out the brackets on the right side will give: x²+3x-2x-6
Tidying this up gives x²+x-6.

e)Now we should bring all the terms to one side to make the equation equal zero.
This gives: x²-3x-28=0

f)We now have a number of methods for solving this equation. We can complete the square, factorise or use the quadratic equation. In this example I will factorise this equation.

This will leave (x+4)(x-7)=0

g)This leaves 2 answers: x=-4 and x=+7

If 3x^2 = 48 find x

Related Maths GCSE answers

5, 11, 21, 35, 53, ... Find the nth term of this sequence.

Jo wants to work out the solutions of x^2 + 3x – 5 = 0. Can the solution be worked out?

(6/x-2)-(2/x+3)=1

Solve 5x-2>3x+11