Solve the simultaneous equations: (1) x^2 + y^2 = 25 and (2) y - 3x = 13

Sub (2) y = 13 + 3x into (1)x^2 + (13 + 3x)^2 = 25x^2 + 169 + 39x + 39x + 9x^2 = 2510x^2 + 78x + 144 = 05x^2 + 39x + 72 = 0 (/2)572 = 360 - need ab=360 such that a+b=39a = 24 b = 155x^2 + 24x + 15x + 72 = 0x(5x + 24) + 3(5x + 24) = 0 [take out the gcf, greatest common factor](x + 3)(5x + 24) = 0x = -3x = -24/5y = 13 + 3(-3) = 4y = 13 + 3(-24/5) = -7/5

MM
Answered by Meghna M. Maths tutor

2808 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Jay, Shelia & Harry share £7200 in the ratio 1:2:5. How much does Harry receive?


Define x and y if 3x+y=11 and 5x-2y=11


Solve these simultaneous equations: y=3x-10; y=2x+5


Solve simultaneously: 2x + y = 18 and x - y = 6


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