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

Start to solve by substitution: eqn 1 x^2 + y^2 = 25eqn 2 y - 3x = 13 => y = 3x + 13Substitute eqn 2 into 1: x^2 + (3x +13)^2 = 25expand and simplify the equation ...5x^2 + 39x + 72 = 0Factorise the equation: (5x+24)(x+3) = 05x = -24 => x = -24/5x = -3Substitute back into equation 2 to find equivalent y values: x = -3 and y = 4, x = -24/5 and y = -7/5

RR

Related Maths GCSE answers

All answers ▸

Expand (x+2)(x-3)(x+4)^2


Solve 3x^2 + 13x + 14 = 0


Expand the brackets: (3a+3)(a+4)


Expand and simplify the following equation: 6(x-3) - 4(x-5) = 0