Solve the simultaneous equations: x^2 + y^2 = 29 and y - x = 3

To solve these equations we need to eliminate one of the variables, so make y the subject of the second equation: y = x + 3. Now y can be substituted into the first equation: x^2 + (x+3)^2 = 29. Expanding (x+3)^2 = x^2 + 6x + 9, so the equation is now x^2 + x^2 + 6x + 9 = 29

Simplifying the equation gives this: 2x^2 + 6x - 20 = 0, which can be simplified further by dividing through by 2: x^2 + 3x - 10 = 0. Factorising the equation gives (x-2)(x+5) = 0, so x = 2 or -5.

Substitute the values of x into y = x+3, to give y = 5 when x = 2 and y = -2 when x = -5.

TG
Answered by Trisha G. Maths tutor

3708 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve these pair of simultaneous questions: 3x+2y=17 4x-y=30


Solve the following simultaneous equations for x and y. 2x+5y=9 and 4x-3y=7


Write x^2+6x+14 in the form of (x+a)^2+b where a and b are constants to be determined.


What is an inverse function?


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