Solve the simultaneous equations: x^2 + y^2 = 29 and y - x = 3

To solve these equations we need to eliminate one of the variables, so make y the subject of the second equation: y = x + 3. Now y can be substituted into the first equation: x^2 + (x+3)^2 = 29. Expanding (x+3)^2 = x^2 + 6x + 9, so the equation is now x^2 + x^2 + 6x + 9 = 29

Simplifying the equation gives this: 2x^2 + 6x - 20 = 0, which can be simplified further by dividing through by 2: x^2 + 3x - 10 = 0. Factorising the equation gives (x-2)(x+5) = 0, so x = 2 or -5.

Substitute the values of x into y = x+3, to give y = 5 when x = 2 and y = -2 when x = -5.

TG
Answered by Trisha G. Maths tutor

3779 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How do you find the length of the diagonal of a rectangle of 5cm by 12cm?


A ladder 5.5m long is leaning against a wall. the foot of the ladder is 1.7m away from the wall. how far up the wall does the ladder reach?


Adam buys 4kg of sweets and pays £10 for them. Adam puts all of the sweers into bags with 250g in each bag. He sells the bags for 70p each. All the bags of sweets are sold, what is the percentage profit?


c is a positive integer. Prove that (6c^3 + 30c) / (3c^2 +15) is an even number.


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