Question

solve the silmutaneous question 3x+y=2 and 3x^2 +y^2+xy=6

Answer 1

ick
solve the first one for y and then substitute in the second one

Answer 2

\[y=2-3x\]
\[3x^2 +y^2+xy=6
\]
\[3x^2+(2-3x)^2+x(2-3x)=6\]

Answer 3

bunch of algebra gives the left hand side as
\[9x^2-10x+4\] so you have
\[9x^2-10x+4=6\] or
\[9x^2-10x-2=0\] quadratic equation is all that is left to do