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

4163 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

We have 2 spinners: spinner A and spinner B. Spinner A can land on 2, 3, 5 or 7. Spinner B can land on 2, 3, 4, 5 or 6. Spin both. Win if one spinner lands on odd and the other lands on even. Play game twice, what is the probability of winning both games?


Simplify. (x(^2)+4x)/(x(^2)+3x-4)


Whats the difference between the three main trigonometric functions?


Solve for x, 4x+3.5 = 1


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