Answer
$x=\frac{2y\pm\sqrt {31y^2+9}}{9}$
$y=\frac{-2x\pm\sqrt {13x^2-1}}{3}$
Work Step by Step
Step 1. Rearrange the equation as a quadratic with $x$ as the variable: $9x^2-(4y)x-(3y^2+1)=0$
Step 2. Let $a=9, b=-4y, c=-(3y^2+1)$ and use the quadratic formula, we have:
$x=\frac{4y\pm\sqrt {16y^2+4(9)(3y^2+1)}}{2(9)}=\frac{2y\pm\sqrt {31y^2+9}}{9}$
Step 3. Rearrange the equation as a quadratic with $y$ as the variable: $3y^2+(4x)y-(9x^2-1)=0$
Step 4. Let $a=3, b=4x, c=-(9x^2-1)$ and use the quadratic formula, we have:
$y=\frac{-4x\pm\sqrt {16x^2+4(9x^2-1)}}{2(3)}=\frac{-2x\pm\sqrt {13x^2-1}}{3}$
