Solve the simultaneous equations x + y = 3 and x^2 + y^2 = 5

Find an expression for y
y = 3 - x
sub this into the second equation
x^2 + (3 - x)^2 = 5
then expand the brackets
x^2 + x^2 - 6x + 9 = 5
simplify
2(x^2 - 3x + 2) = 0
factorise
2(x - 2)(x - 1) = 0
so x = 2, 1
to find the y value sub into the original equation to get
y = 3 - 1 = 2
y = 3 - 2 = 1
so the solutions are
y = 2, x = 1
and
y = 1, x = 2

KW
Answered by Kate W. Maths tutor

15882 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

f(x) = 3x - 2a || g(x) = 2ax + 1 || fg(x) = 2x + b/2


Find the equation of the line perpendicular to y=2x-1 that passes through (2,0)


Write x^2 + 4x - 16 in the form (x+a)^2-b


Factorise 2(x^2) +7x+3


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