Question 867540
Some help is this:


(a) gives you y=2x-3.  Substitute in (b) and you have,
{{{x^2+2x(2x-3)=1}}}
simplifying to...
{{{x^2+4x^2-6x=1}}}
{{{5x^2-6x-1=0}}}.
Can you factor this and find x values?  If not factorable, you can use the general solution of a quadratic equation and find the x values.  Use them to get the corresponding y values.



(5x 1)(x 1)...   NOT factorable.
{{{x=(6+- sqrt((-6)^2-4*5*(-1)))/(2*5)}}}