Solve algebraically the simultaneous equations: x^2 + y^2 = 25 and y - 3x = 13

Firstly, we need to use one equation to find an expression for one variable in terms of the other. Then we can substitute this expression into the other equation and solve for that variable. Using this numerical value we can then solve for the other variable using either of the original equations. Using y - 3x = 13 we can obtain y = 13 + 3x Then substitute this in for y in the other equation: x² + (13 + 3x)² = 25 Expand and simplify: x² + 169 + 39x + 39x + 9x² = 25; 10x² + 78x + 169 = 25; 10x² + 78x + 144 = 0 We can divide both sides by 2 to simplify further: 5x² + 39x + 72 = 0 Now we have a quadratic equation to solve. To solve by factorisation, we need to think of two numbers which multiply to 5 x 72 (=360) and add to 39. Those numbers are 24 and 15. We can then rewrite the quadratic equation as: 5x² + 15x + 24x + 72 = 0 Then factorise. 5x(x+3) + 24(x+3) = 0; (5x+24)(x+3) = 0 So either 5x + 24 = 0 or x + 3 = 0 So x = -24/5 or x = -3Substitute each of these values of x into either original equation to find the corresponding value of y: When x = -24/5, y = 13 + 3x, y = 13 - 72/5, y = 65/5 - 72/5, y = -7/5; When x = -3, y = 13 -9, y = 4