Step 1: Recognise the quadratic term, this cannot be solved through elimination. Therefore we must need to use substitution. Substitute y = x + 3 into the quadratic. Gives x + 3 = x2 + 3xStep 2: Rearrange to get all terms on one side (the form of a familiar quadratic). Gives x2 + 2x - 3 = 0Step 3: Factorise. Gives (x + 3)(x - 1) = 0. Step 4: If a product of two numbers equals zero, one or both of the numbers must be zero. Set each bracket equal to zero to obtain solutions of x. Gives x = -3, 1Step 5: Sub each solution for x back into y = x + 3 to give y = 0, 4. Solutions are x = -3, y = 0 and x = 1, y = 4.Extension: This question can be easily extended by asking the student to solve the simultaneous equations graphically (tests ability to draw straight lines and quadratics).