Answer
$(\frac{60}{13}, \pm\frac{60}{13})$, $(-\frac{60}{13}, \pm\frac{60}{13})$
Work Step by Step
Step 1. Subtract the first equation from the second, we get $119(x^2-y^2)=0$, thus $y^2=x^2$ or $y=\pm x$
Step 2. The first equation becomes $169x^2=3600$ so that $x=\pm\frac{60}{13}$
Step 3. State the 4 intersection points: $(\frac{60}{13}, \pm\frac{60}{13})$, $(-\frac{60}{13}, \pm\frac{60}{13})$
Step 4. Graph the equations and label the intersection points as shown.
