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

To solve these equations we need to eliminate one of the variables, so make y the subject of the second equation: y = x + 3. Now y can be substituted into the first equation: x^2 + (x+3)^2 = 29. Expanding (x+3)^2 = x^2 + 6x + 9, so the equation is now x^2 + x^2 + 6x + 9 = 29

Simplifying the equation gives this: 2x^2 + 6x - 20 = 0, which can be simplified further by dividing through by 2: x^2 + 3x - 10 = 0. Factorising the equation gives (x-2)(x+5) = 0, so x = 2 or -5.

Substitute the values of x into y = x+3, to give y = 5 when x = 2 and y = -2 when x = -5.

TG

Related Maths GCSE answers

All answers ▸

solve to 2 decimal places; (2x+3/(x-4)) - 2x-8/(2x+1) = 1


Solve the following pair of simultaneous equations: 1. 3x + 2y = 9 2. 6x + 5y = 21


Simplify: y+y-x×x


1