Answer
$x=\frac{3y\pm\sqrt {5-y^2}}{5}$
$y=\frac{3x\pm\sqrt {2-x^2}}{2}$
Work Step by Step
Step 1. Rearrange the equation as a quadratic with $x$ as the variable: $5x^2-(6y)x+(2y^2-1)=0$
Step 2. Let $a=5, b=-6y, c=(2y^2-1)$ and use the quadratic formula, we have:
$x=\frac{6y\pm\sqrt {36y^2-4(5)(2y^2-1)}}{2(5)}=\frac{3y\pm\sqrt {5-y^2}}{5}$
Step 3. Rearrange the equation as a quadratic with $y$ as the variable: $2y^2-(6x)y+(5x^2-1)=0$
Step 4. Let $a=2, b=-6x, c=(5x^2-1)$ and use the quadratic formula, we have:
$y=\frac{6x\pm\sqrt {36x^2-4(2)(5x^2-1)}}{2(2)}=\frac{3x\pm\sqrt {2-x^2}}{2}$
