How do I solve simultaneous equations given a linear and a quadratic equation?

Let's use the following example: y - x = 3 y^2 + x^2 = 29 . Firstly, the second equation is the quadratic equation because it has a y-squared term. The first equation is linear because it doesn't have any squared x or y terms. Now, take the linear equation and rewrite it as y = x + 3 (by adding x to both sides of the equation). Then substitute this result into the quadratic equation. What this means is: wherever you see a y in the second equation, instead you write x + 3. This gives: (x + 3)^2 + x^2 = 29 . To simplify this, multiply out the brackets using the following steps: (x + 3)(x + 3) + x^2 = 29, x^2 + 3x + 3x + 9 + x^2 = 29, 2x^2 + 6x - 20 = 0, x^2 + 3x - 10 = 0 . At this point you have a quadratic equation. If you are confident factorising it, this will be the quickest method. Otherwise, you could try using the quadratic formula. You get the following two solutions: x = -5 x = 2 . Substitute these two answers back into the first (the linear) equation. This gives the following solutions: when x = -5, y = -2; when x = 2, y = 5.

Answered by Erica H. Maths tutor

4480 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

The area of this rectangle is 56 cm2 length = 3k+2 and width = 7 - Find the value of k


Solve the simultaneous equations 2x + 3y = 6 - 3x and 5x + 6y = 10 - y.


a) A line passes through (0,9) and (3,12) write down the equation of this line . b) A line perpendicular to the line in part a passes through the point (3,14) write the equation of this line.)


How do you factorise x^2 -4 = 0?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences