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

This question is done by substituting either the x or y. To do this, rearrange the second equation to make y the subject. We make y the subject as it keeps the equation and calculation simple. After rearranging the equation, we get y=13+3x. Now, sub this into the first equation to get x^2+(13+3x)^2=25. Expand the brackets out first to get x^2+169+6x+9x^2=25. Then add or subtract (in this case only add) the same x's to get 10x^2+6x+169=25. Now bring the 25 over to the left to get 10x^2+6x+144=0. Divide through by 2 to simplify to get 5x^2+3x+72=0. Then factorise to get (x+3)(5x+24)=0. Then separately equal the two brackets with 0. Then X are x=-3, x=-(24/5). Then sub these back into the equation given by the question. This case y-3x=13 as its easier to calculate. So y=4, y=-(7/5).

SS
Answered by Shun S. Maths tutor

21145 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Mixed rugby team of 20, 5 are female. 15 play at a time. i.) What is the percentage chance of a female playing. ii.)A minimum of three females must now be on the pitch. What is the percentage chance of 4 females playing?


How do I solve this pair of simultaneous equations: 3x+y=7 and 3x-y=5?


For the equation y=2x+7, what is the gradient and the y intercept?


How to differentiate 9x^2+ 4x-7=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