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

First begin with rearranging the second equation to put it in terms of Y. Then substitute by putting the second equation (y=3x+13) into equation one as Y^2. Expand (3x+13)^2 and add like terms to form 10x^2 + 78x + 144 =0. Simplify and fractionise to get the terms of x=-3 and x= -24/5. Finally substitute each term of X into the original equation of (y=3x+13) to get the Y terms of Y=4 and Y=-3/5

DL

Related Maths GCSE answers

All answers ▸

Rearrange the following to make c the subject: 11a + 5c = d ( 6 + 2c )


How to factorise a simple linear equation such as '9Y + 6'


Express 60 as a product of its prime factors.


5q^2.p^12/10(q.p^3)^2