Question 331485
lol anyone but john...lol he must of gave u a bad answer..lol
{{{2x-y=1}}}
{{{2x=1+y}}}
{{{(2x-1)=y}}} substitute this in other equation
<p>
{{{x(2x-1)=1}}}
{{{2x^2-x=1}}}
{{{2x^2-x-1=0}}} View quadratic equation below
x=1,-0.5
to find y simply plug in the x's
y=1,0

*[invoke quadratic "x", 2, -1, -1]