Solve the simultaneous equations: (1) x^2 + y^2=41 and (2) y=2x-3

First we substitute one x or y into the other equation. The easiest one to put in in this case would be the y in equation (2). So putting the y into (1) you get x^2 + (2x-3)^2=41. Then we expand out the brackets using FOIL: 5x^2-12x-32=0. we then would factorise this by finding two numbers that times to give 160 and add to give -12. these would be -20 and 8. Because we have a 5x^2 this would factorise to give (5x+8)(x-4)=0. The solutions of these would be x=-8/5 and x=4, and then substituting the xs back into one equation (this can be either) to give y=-31/5 and y=5.