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

1503 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Expand and simplify (2x + 5) (3x - 2)


A bag has 3 red balls and 5 green balls. I take out 2 balls, without replacing them. What is the probability of choosing at least one red ball? Give your answer to 3 decimal places.


Solve the simultaneous equations x^2 +8y=20 and y=x+4


Integrate y=3x+2 with respect to x between the bounds x=5 and x=0, and state the physical significance of this.


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 2013–2024

Terms & Conditions|Privacy Policy