How to solve simultaneous equations with a quadratic

Example.

" Solve the simultaneous equations

y + 4x + 1 = 0

y^2 + 5x^2 + 2x = 0 "

We can rearrange the first equation to something that we can substitute into the second equation, for example

y = -4x - 1

We then substitute this into the second equation

(-4x - 1)^2 + 5x^2 + 2x = 0

16x^2 + 8x + 1 + 5x^2 + 2x = 0

21x^2 + 10x + 1 = 0

(7x + 1)(3x + 1) = 0

Gives us the solutions

x = -1/7 and x = -1/3

We then put these values into our first equation (y = -4x - 1) to give

y = -3/7 and y = 1/3

TW
Answered by Tilly W. Maths tutor

2961 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Consider the function f (x) = (2/3) x^3 + bx^2 + 2x + 3, where b is some undetermined coefficient: (a) find f'(x) and f''(x) and (b) if you know that f(x) has a stationary point at x = 2, use this information to find b.


How can I find the normal to a curve at a given point?


Given that 5cos^2(x) - cos(x) = sin^2(x), find the possible values of cos(x) using a suitable quadratic equation.


A curve is defined by the parametric equations x=t^2/2 +1 and y=4/t -1. Find the gradient of the curve at t=2 and an equation for the curve in terms of just x and y.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences