x+y=2 (1) x2+2y=12 (2) As we are dealing with x2 we’re going to have to use substitution to solve this set of simultaneous equations. From the first equation, we can rearrange to make y the subject: y=2-x (3)Now, we can substitute our new equation (3) into equation (2) to eliminate y: x2 + 2(2-x) = 12. By expanding we get: x2 + 4-2x = 12. To solve for x we need to make the equation equal to 0: x2-2x-8=0. To factorise, we need 2 numbers that multiply to make -8 and add to make -2; these 2 numbers are -4 and 2. Therefore x2-2x-8=0 is equal to (x-4)(x+2)=0. Solving for x we get x=4 and x=-2. We can now substitute these into our equation (3) to find y: y=2-4=-2 and y=2-(-2)=4. So our 2 pairs of answers are: x=4, y=-2 and x=-2, y=4