Solve the simultaneous equations: y = x + 6, x^2 + 2y = 9

y = x + 3x2 + 2y = 9We want to make the squared equation have either only x's or only y's, so we substitute y = x + 3 into the equation:x2 + 2(x + 3) = 9We need to take out the brackets, and simplify this equation.x2 + 2x + 6 = 9x2 + 2x - 3 = 0This is now in the form of a quadratic, so we can solve it:x2 + 2x - 3 = 0(x + 3)(x - 1) = 0 so therefore, x = -3, x = 1These answers for the x-values can give us the y-values, if we substitute them into the original equation:When x = -3, y = -3 + 3 y = 0When x = 1 y = 1 + 3 y = 4We can check our answers by substituting them into the original equation:x = -3, y = 0] (3)2 + 2(0) = 9 = CORRECTx = 1, y = 4] (1)2 + 2(4) = 1 + 8 = 9 = CORRECT

HH
Answered by Hayley H. Maths tutor

3195 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

solve the simultaneous equation x^2 + 2y = 9 , y - x = 3


Solve the simultaneous equations x^2 + y^2 =13 and x= y - 5.


Solve the simultaneous equations: 6x + 2y = -3, 4x - 3y = 11


Find the values of k for which the equation (2k-3)x^2- kx+(k-1)=0 has equal roots.


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