2y-3x=2 & 4y2-4xy=18x2=5
It is not very clear. Did you mean?:
$\displaystyle \begin{Bmatrix} 2y-3x=2\\4y^2-4xy-18x^2=5\end{matrix}$
Fernando Revilla
All right, I've corrected it. Then from the first equation:
$\displaystyle y=\dfrac{2+3x}{2}$
Now, substitute $\displaystyle y$ in the second one and you'll obtain a quadratic equation on $\displaystyle x$ .
Fernando Revilla
Please show some effort. See rule #11: http://www.mathhelpforum.com/math-he...ng-151418.html.
Post what you can do and make it clear where you get stuck.