Solve these simultaneous equations. x^2+y^2=9, x-y+3=0
Rearrage so you have x+3=y, and x2 +y2 =9. Substitute the equation into another so you have x2+(x+3)2=9. Expand the brackets so that you have x2+x2+6x+9=9. Then you rearrange so it's all on one side of the equation and equal to zero you will then have 2x2+6x=0 you can the divide by two so you have x2+3x=0. You then have to factorise so you have x(x+3)=0. This means you now know the value of x, it is either -3 or 0. You then substitute those values int either one of the two equations (usually the easier one) so you now have y=3 when x=0 and y=0 when x=-3.
AB
Related Maths GCSE answers
All answers ▸A particle P of mass 0.4 kg is moving under the action of a constant force F newtons. Initially the velocity of P is (6i – 27j) m s−1 and 4 s later the velocity of P is (−14i + 21j) m s−1 . Find, in terms of i and j, the acceleration of P.
Using factorization, solve x^2 + 10x + 24 = 0
Solve the following simultaneous equations: 4x+y=10 and 2x-3y=19
Solve 5x-6=3x+7