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

Steps:Number both equationsUsing equation 2, solve for y in terms of x.Substitute this answer into equation 1 so that your equation only includes x terms and no y terms.Solve equation 1 for x. (This will be in numerical terms)Substitute the answer for x into equation 2, to find y.E.g.Steps:Number equations:Equation 1: x^2 + y^2 = 25Equation 2: y - 3x = 13Solve y - 3x = 13 in terms of xy = 13 + 3xSubstitute in 'y = 13 + 3x' into equation 1: x^2 + y^2 = 25x^2 + (13 + 3x)^2 = 25Solve the new equation 1 by multiplying out the brackets and solving for x.x^2 + (13 +3x)(13+ 3x) = 25x^2 + 169 + 9x^2 + 39x + 39x = 2510x^2 + 78x + 144 = 0Then divide equation by 2 to get it in it's simplest form and factorise, giving you: (5x + 24)(x + 3) = 0Hence, x = -3Substitute x = -3 into y - 3x = 13:y - 3(-3) = 13y = 13 - 9Answer:y = 4x = -3