How do you solve simultaneous equations?

To solve simultaneous equations you need to cancel out variables like x and y one at a time to solve for the other. In the edexcel GCSE 2017 paper this question came up: solve simultaneously x2+y2=25 y-3x=13
From this you can see it's easier to work with the second equation first. You can move this around to make y=3x+13, then substitute this into the first equation to get: x2+(3x+13)2=25. You should then expand this bracket out and find x2+9x2+39x+39x+169=25 and condense this to 10x2+78x+144=0. It can then be divided by 2 to get 5x2+39x+72=0
This then needs to be factorised as: (x+3)(5x+24)=0. You then must set each individual bracket equal to 0 and obtain x=-3 and x=-24/5. Now you have found these you can easily substitute them into our y=3x+13 equation and find that y=3(-3)+13 which gives y=4 and y=3(-24/5)+13 which gives y=-7/5SOLVED

LN
Answered by Libby N. Maths tutor

2917 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Express x^2+2x+4 in the form (x+a)^2 + b, hence find the stationary point of the curve.


Solve the following simultaneous equations: x^2 + 2y = 9, y = x + 3


Solve ((6+x)/2) + ((2-3x)/3) = 31/6


Dipen and Nisha are planning their wedding reception. Nisha says, “I want to invite 70 guests.” Dipen says, “If we invite one-fifth fewer guests, we will save more than £500” Is Dipen correct? (Taken from Nov 2014 AQA Unit 2)


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:

© 2025 by IXL Learning