Solve the simultaneous equations: x^2 + y^2 = 9 and x + y = 2

Firstly we need to realise that equation 1 (x^2 + y^2 = 9) does not factorise. So once realising this, we must rearrange equation 2 (x + y = 2) so that y = 2 - x. Now we substitute (y = 2 - x) into equation 1, so that we have a new equation, equation 3: x^2 + (2 - x)^2 = 9. Now we must multiply out the brackets this giving, x^2 + 4 - 2x - 2x + x^2 = 9. Now we simplify this and rearrange so we have all the terms in the equation equal to 0 (quadratic form): 2x^2 - 4x - 5 = 0. We must now use the quadratic formula since this we cannot factorise this equation:
ax^2 + bx + c = 0x = (-b +/- sqrt(b^2 - 4ac))/ 2ab(show using the whiteboard)
We now use this formula to find x. We know that a = 2, b = -4 and c = -5Now we plug these numbers into the quadratic formula, this will lead to x = (4 +/- sqrt(56)) / 4. Note the +/-, this will mean that we will get two answers for x. We get the answers x = 2.87 or x = -0.87(2dp). Now for each of these answers, plug them into the rearranged equation 2 ( y = 2 - x). This will lead to the answers: y = -0.87 or y = 2.87 (2dp). Now we have finished the question, when writing the answer however don't forget to group the answers.The 2 pairs of answers: x = 2.87 and y = -0.87 or x = -0.87 and y = 2.87