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

4031 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Show that the following 2 lines are parallel: l1: 3y=15x+17 l2: 7y+5=35x


A cuboid has edge 7 centimetres, 5 centimetres and a total surface area of 142 centimetres squared. A larger, similar cuboid has a shortest edge of 12 centimetres. Find the third edge of the smaller cuboid and the volume of the larger cuboid.


Given 6x+2y=4 and 5x+y=8, solve the simultaneous equations to find x and y.


How do I find the length of a side of a triangle using the cosine rule?


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