Solve algebraically the simultaneous equations x^2 +y^2 =25, y – 3x = 13

This question is done by substituting either the x or y. To do this, rearrange the second equation to make y the subject. We make y the subject as it keeps the equation and calculation simple. After rearranging the equation, we get y=13+3x. Now, sub this into the first equation to get x^2+(13+3x)^2=25. Expand the brackets out first to get x^2+169+6x+9x^2=25. Then add or subtract (in this case only add) the same x's to get 10x^2+6x+169=25. Now bring the 25 over to the left to get 10x^2+6x+144=0. Divide through by 2 to simplify to get 5x^2+3x+72=0. Then factorise to get (x+3)(5x+24)=0. Then separately equal the two brackets with 0. Then X are x=-3, x=-(24/5). Then sub these back into the equation given by the question. This case y-3x=13 as its easier to calculate. So y=4, y=-(7/5).