Answer
The possible combination of solutions is.
$(1,3),(-1,3),(1,-3)$ and $(-1,-3)$.
Work Step by Step
The given equations are
$3x^2+4y^2=39$ ... (1)
$5x^2-2y^2=-13$ ... (2)
Multiply equation (2) by $2$.
$10x^2-4y^2=-26$ ... (3)
Add equation (1) and (3).
$3x^2+4y^2+10x^2-4y^2=39-26$
$13x^2=13$
$x^2=1$
$x=\pm 1$. Substitute into equation (1).
$3(\pm 1)^2+4y^2=39$
$3+4y^2=39$
$4y^2=39-3$
$4y^2=36$
$y^2=\frac{36}{4}$
$y^2=9$
$y=\pm 3$.
