Question

I need help solving this system: 7x^2-9y^2=31 and 5x^2+9y^2=161. It would be nice to have an explanation too.

Answer 1

add them as a start. the y^2 will drop out and you will have a quadratic in x

Answer 2

ok so we have:

\[7x ^{2}-9y ^{2}=31\]
\[5x ^{2}+9y ^{2}=161\]


add and we get:\[12x ^{2}=192\]

divide by 12

\[x^2=16\]
\[x=4\]

plug in to find y

\[7(4)^2-9y^2=31\]

\[112-9y^2=31\]

\[-9y^2=-81\]

\[y^2= 9\]

\[y=3\]