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

Answered by Esther M. Maths tutor

2070 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How do you solve these simultaneous equations? 4x + 5y = 8; 2x + 3y = 5


How can you expand brackets? e.g: (x-4)(x+7)


How do I rearrange and make y the subject in equations such as "(y/4) - X = 1"?


Differentiate: 6x^2 + 5x +7 =y


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences