GCSE 2011: Solve the simultaneous equations: y^2 = 2x + 29 and y = x - 3

This was the final question in AQA's 2011 non-calculator paper, and most top candidates will be looking to score all 5 marks that are on offer. Our usual technique for simultaneous equations isn't going to work , because we have a y^2 rather than a y. Instead we'll need to square the 2nd equation: y = x - 3, y^2 = (x - 3)^2 = x^2 - 6x + 9 From the first equation we have: y^2 = 2x + 29. Setting these two equations equal gives: 2x + 29 = x^2 - 6x + 9. Subtracting (2x + 29) from both sides gives: 0 = x^2 - 8x - 20. Factorising we get: 0 = (x-10)(x+2). We could use the quadratic formula or completing the square if you prefer. x = 10, x = -2. We've got our values of x; just need to sub back into y = x - 3 to get our answer: x = 10 implies y = 10 - 3 = 7. x = -2 implies y = -2 - 3 = -5. We could put these back into the first equation to check we have the right answer. Not too bad for the very last question, although care is needed making the substitution to get rid of y.