Answer
$(\pm \sqrt{\frac{25}{2}},-3)$ and $(\pm \sqrt{\frac{9}{2}},1)$
Work Step by Step
Step 1. Multiply Equation 2 by 2
$2x^2+4y=13$
$2x^2-2y^2=7$
Step 2. Substract both equations and solve for $y$
$2y^2+4y=6$ Divide by 2
$y^2+2y=3$
$y^2+2y-3=0$ Factorize
$(y+3)(y-1)=0$
$y=-3$ and $y=1$
Step 3. Back substitute to solve for $x$
For $y=-3$,
$2x^2+4\cdot (-3)=13$
$2x^2=25$
$x^2=\frac{25}{2}$
$x=\pm \sqrt{\frac{25}{2}}$
For $y=1$,
$2x^2+4\cdot 1=13$
$2x^2=9$
$x^2=\frac{9}{2}$
$x=\pm \sqrt{\frac{9}{2}}$
Step 4. Conclude the solutions
The solutions are $(\pm \sqrt{\frac{25}{2}},-3)$ and $(\pm \sqrt{\frac{9}{2}},1)$.
