Solve algebraically the simultaneous equations, x^2 + y^2 = 25 and y – 3x = 13

First you need to pick a variable to solve for( irrelevant which one is picked) so I will choose to solve y first as in this question it is easier (as y has no coefficient). So make y the subject of the simpler equation(without the power) to get y = 13 + 3x, then sub this into the first equation to eliminate the y variable to get x^2 + (13 + 3x)^2 = 25. Then simplify and group terms (also putting all terms on the same side) to get quadratic in x: 10x^2 + 78x + 144 = 0. Then can simplify by factoring out 2 to get 5x^2 + 39x + 72 = 0. Then use quadratic formula or factorising into (5x + 2)(x + 3) get solutions for x being -3 and -24/5. Then can sub these into the original equation to find solutions for y being 4 and -7/5 respectively. So solutions are x = -3, y = 4 and x = -24/5 and y is -7/5.

BU
Answered by Benjamin U. Maths tutor

2970 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve the simultaneous equations 5x + 2y = 4 and x - y - 5 =0


work out 3 1/2 - 2 1/3. Give answer as an improper fraction


If a right angled triangle has its longest side 7cm and another side is 4cm then how long is the other side of the triangle? Show your working


Fully simplify (8a^2b * ab^4)/(2a^3b^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:

© 2026 by IXL Learning