Solve the two simultaneous equations. 1. x^2 + y^2 = 25, 2. y - 3x = 13

The way to solve simultaneous equations is to 'slot in' one equation into the other. For this question, we will use the information in equation 2 (after manipulating it) to solve equation one. First, we add '3x' to both sides of equation 2. This results in y = 13 + 3x. If we were to square both sides, this would lead us to get y^2 = (13+3x)(13+3x). We then substitute 'y^2' in equation 1 to get x^2 + (13+3x)(13+3x) = 25. We can then expand the brackets to get x^2 + 9x^2 + 78x + 169 = 25. If we add the common terms, we get 10x^2 + 78x + 169 = 25. Then we can subtract 25 from both sides which results in 10x^2 + 78x + 144 = 0. We can then half both sides (to make it easier to work with) which is 5x^2 + 39x + 72 = 0. Then we can separate the 'x' terms to get two different 'x' terms. We want one to be a multiple of 5 (since the coefficient of x^2 is 5). We can then get 5x^2 + 15x + 24x + 72 = 0. Now we can factorise this equation to 5x(x+3) + 24(x+3) = 0. Finally we can factories this fully into (5x+24)(x+3). The results of this is 5x = -24, x = -24/5. The second value of x we get is x = -3. If we then use these x values for the original second equation we get y - 3(-24/5) = 13. y + 72/5 = 13.(we then covert 13 into a division of 5 to get common terms. y + 72/5 = 65/5. Then y = -7/5. For the second value of x we use simple manipulation y - 3(-3) = 13. y + 9 = 13. y =4. Our four final values are x = -24/5, y = -7/5, x = -3, y =4.