How do I solve simultaneous equations that aren't linear, for example x^2 + 2y = 9, y = x + 3

First, let's start by labelling the equations. We can call x2+2y=9 equation 1 and y=x+3 equation 2. Rearrange equation 2 to give us x = y-3. We can then substitute this back into equation 1. So we get (y-3)2+ 2y = 9Expanding these brackets gives y2 - 4y = 0 . We can factorise here to give y(y-4)=0 so we have 2 cases, case 1: y=0, or case 2: (y-4) = 0, so y=4. By substituting y=0 back into equation 1, we can see that x=-3, By substituting y=4 back into equation 1, we get x=1. So these are our solutions, either we have x=-3, y=0 or we have that x=1, y=4

EM

Related Maths GCSE answers

All answers ▸

Solve: x^2 + 2x - 3 = 0


Work out the value of 2^14 ÷ (2^9)^2


Sketching a quadratic


On a packet of brown rice it says 'When 60g of rice is cooked, it will weigh 145g.' If Katy has 100g of brown rice, how much will it weigh when cooked?'