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

This question is slightly trickier than normal simultaneous equations, because we have values to the power of 2. What we can do is make either y or x the subject of the 2nd equation (y-x = 3). For this example, I will choose to make y the subject, which then gives us y = 3 + x. We can use this equation, and substitute it into the value of "y" in the 1st equation, as follows --> x^2 + (3+x)^2 = 29 We can then expand the brackets and simplify: x^2 + x^2 + 6x + 9 = 29 2x^2 + 6x -20 = 0 (it is important to make the equation equal to 0 so that we can solve the equation to find the values of x) x^2 + 3x - 10 = 0 (we can divide the whole equation by 2, as this is a common factor) (x + 5) (x - 2) = 0 (we can factorise the equation to give us 2 brackets; we have found two numbers which multiply to give -10 and add to give +3) .˙. x = -5 and x = 2 (each bracket is made equal to 0 and solved separately, we have two values for x because this is a quadratic equation)Each value is then substituted back into the rearranged 2nd equation (y = 3 + x) which gives us y = -2 and y = 5