Solve the simultaneous equations x^2+ y^2 = 29 and y–x = 3

For this question we will be using the substitution method in order to solve this question. Firstly we will take the less complex equation and rearrange to make either x or y the subject of the formula. We will make y the subject of the formula in this case. So y-x = 3 becomes y = x+3. We will now substitute this equation for y into our other equation. So x^2+y^2 = 29 will now become x^2+(x+3)^2 = 29. Next we will expand and simplify this equation. x^2+(x+3)^2 = 29 now becomes x^2+3x-10=0. This is a quadratic equation, this can be solved using the quadratic formula giving answers of x = -5 and x = 2. These two values for x can now be substituted back into the eqaution y-x = 3 to give y = -2 (for x = -5) and y = 5 (for x=2). These are the solutions to the equation.