Solve algebraically the simultaneous equations x^2 + y^2 = 25 and y - x = 1
Let's label the equations x2 + y2 = 25 (1) y - x = 1 (2). We can use substitution to solve this simultaneous equation. Let's make y the subject of the equation (2) and we get y = x + 1. Now, lets find y2 and get y2 = (x + 1)2 = (x + 1)(x + 1) = x2 + 2x + 1. Now we can substitute y2 into equation (1) and get the following:x2 + x2 + 2x + 1 = 25 => 2x2 + 2x -24 = 0 => x2 + x - 12 = 0. Factorising gives us (x + 4)(x - 3) = 0 and the roots are x = -4 and x = 3Now we need to find y by substituting x into y = x + 1 and we get y = -3 and y = 4. Therefore, the solutions are x = -4, y = -3 and x = 3, y = 4.
