Solve algebraically the simultaneous equations x^2 +y^2 = 25, y – 3x = 13

  1. Rearrange y - 3x = 13y = 3x + 13 2. Substitute into x^2 +y^2 = 25x^2 + (3x + 13)^2 = 25 3. Expand bracketx^2 +(9x^2 + 39x +39x +169) = 25 4. Form quadratic equation 10x^2 + 78x + 144 = 0 (Dividing by 2 makes factorising easier)5x^2 + 39x + 72 = 0 5. Factorise (5x + 24)(x + 3) = 0 6. Solve for xx = -24/5, x = -3 7. Substitute values of x into y = 3x +13 to find y values x = -24/5, y = -7/5x = -3, y = 4
EN
Answered by Emily N. Maths tutor

4695 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Three identical isosceles triangles are joined together to make a trapezium. Each triangle has base b cm and height h cm. Work out an expression, in terms of b and h for the area of the trapezium.


Solve the following simultaneous equations. x^2+y^2=25. y-3x=13


All tickets for a concert are the same price. Ayo and Rachel pay £63 total for some tickets. Ayo pays £24.50 for 7 tickets. How many tickets does Rachel buy?


How to work out a fraction of an amount?


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