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

2704 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve: x/x+4 + 7/x-2 = 1 (Must show working)[4 marks]


How do you work out the tangent to a circle at a given point on the circle?


[Higher] Simplify the following expressions: x^7 X x ^5 and (x^-2)^-3


Solve x^(2) + 7x + 6 = 0.


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences