Answer
$(\sqrt \frac{9}{2}, 1), (- \sqrt \frac{9}{2}, 1), ( \sqrt \frac{25}{2}, 3) and (- \sqrt \frac{25}{2}, 3) $
Work Step by Step
$2x^2 + 4y = 13$ (1)
$x^2 - y^2 = 7/2$ (2)
(1) - 2 (2)
$4y + 2y^2 = 6$
$2y^2 + 4y - 6 = 0$
$(2y - 2) (y + 3) = 0$
y = 1, -3
Substitute y= 1
$2x^2 + 4(1) = 13$
$x^2 = 9/2$
$x = +/- \sqrt \frac{9}{2}$
Substitute y= -3
$2x^2 + 4(-3) = 13$
$x^2 = 25/2$
$x = +/- \sqrt \frac{25}{2}$
Thus the answers are $(\sqrt \frac{9}{2}, 1), (- \sqrt \frac{9}{2}, 1), ( \sqrt \frac{25}{2}, 3) and (- \sqrt \frac{25}{2}, 3) $
