The line y=2x- 3 intersects the curve y=x^2+x-15 at the points
(a)(3,3)and(4,5)
(b)(-3,-9)and(4,5)
(c)(3,3)and(-4,-11)
(d)(-3,-9)and(-4,-11).
At a point of intersection:Originally Posted by bobby77
$\displaystyle
2x-3=x^2+x-15
$,
so:
$\displaystyle
x^2-x-12=0
$
Now trying the candidates for $\displaystyle x$, we find that ,$\displaystyle x=-3$ satisfies
this equation as does $\displaystyle x=4$, so (b) must be the answer (assuming the
$\displaystyle y$'s are OK which I suppose we can).
RonL