Solve the following simultaneous equation: 1) 2x=y-5, 2) 2y^2=4x^2+4x-15

The most instinctual method one would go by is to rearrange the terms so that one side of the equation will be equal to either x,y, or zero. In this case to avoid fractions it may be best to rearrange the equation in terms of y, hence y=2x+5- the second equation does not need to be rearranged. Examining Eq2, we see that it is possible to factorise is hence getting 2y^2=(2x-3)(2x+5). Substituting Eq 1into Eq2, we get: 2(2x+5)^2=(2x-3)(2x+5). Cancellation of (2x+5) on both sides is possible here, so we are left with: 2(2x+5)=2x-3, which when rearranged to make 'x' the subject gives x= -13/2.