Solve 9x^2+16y^2-36x-96v+36=0

$$9 x ^ { 2 } + 16 y ^ { 2 } - 36 x - 96 v + 36 = 0$$

Solve for v

$$v=\frac{3\left(x-2\right)^{2}}{32}+\frac{y^{2}}{6}$$

Steps for Solving Linear Equation

Subtract 9x^{2} from both sides. Anything subtracted from zero gives its negation.

$$16y^{2}-36x-96v+36=-9x^{2}$$

Subtract 16y^{2} from both sides.

$$-36x-96v+36=-9x^{2}-16y^{2}$$

Add 36x to both sides.

$$-96v+36=-9x^{2}-16y^{2}+36x$$

Subtract 36 from both sides.

$$-96v=-9x^{2}-16y^{2}+36x-36$$

The equation is in standard form.

$$-96v=-9x^{2}+36x-16y^{2}-36$$

Divide both sides by -96.

$$\frac{-96v}{-96}=\frac{-9\left(x-2\right)^{2}-16y^{2}}{-96}$$

Dividing by -96 undoes the multiplication by -96.

$$v=\frac{-9\left(x-2\right)^{2}-16y^{2}}{-96}$$

Divide -16y^{2}-9\left(-2+x\right)^{2} by -96.

$$v=\frac{3\left(x-2\right)^{2}}{32}+\frac{y^{2}}{6}$$

Solve for x (complex solution)

$$x=\frac{4\sqrt{6v-y^{2}}}{3}+2$$

$$x=-\frac{4\sqrt{6v-y^{2}}}{3}+2$$

Solve for x

$$x=\frac{4\sqrt{6v-y^{2}}}{3}+2$$

$$x=-\frac{4\sqrt{6v-y^{2}}}{3}+2\text{, }v\geq \frac{y^{2}}{6}$$