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
Answered by Esther M. Maths tutor

2496 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

write log2(5) +log2​​​​​​​(3) in its simplest form


Expand the expression (3x+2)(3-2x)


Bhavin, Max and Imran share 6000 rupees in the ratios 2 : 3 : 7. Imran then gives 3/5 of his share of the money to Bhavin. What percentage of the 6000 rupees does Bhavin now have? Give your answer correct to the nearest whole number.


How do I remember the area of a circle if it's not in the formula book?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning