Solve these two equations simultaneously: 7x + y = 1 and 2x^2 - y = 3.

Equation 1: 2x2 - y = 3 and Equation 2: 7x + y = 1. Rearrange equation 1 to make y the subject: 2x2 - y = 3, 2x2 - 3 = y. Substitute equation 1 into equation 2 to get equation 3: 7x + (2x2 - 3) = 1, 7x + 2x2 - 3 = 1, 2x2 + 7x - 3 = 1, 2x2 + 7x - 4 = 0. Solve equation 3 to get values of x: 2x2 + 7x - 4 = 0, (2x - 1)(x + 4) = 0 so 2x - 1= 0, 2x = 1, x = 1/2 and x + 4 = 0, x = -4. Substitute the values of x into equation 2: 7x + y = 1, 7(1/2) + y = 1, 7/2 + y = 1, y = -5/2 and 7x + y = 1, 7(-4) + y = 1, -28 +y = 1, y = 29. Therefore the answer to the solution is: x = 1/2 y = -5/2 and x = -4 y = 29.

DD
Answered by Darcie D. Maths tutor

3663 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Show that the recurring decimal 0.13636... can be written as the fraction 3/22


How do I differentiate?


Expand and simplify the following expression: x(5x – 2) – 3(x2 – 2x + 7)


p and q are two numbers each greater than zero. √(p^2 + 5q) = 8 and √(p^2 – 3q) = 6. Find the values of p and q.


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