Solve (3y+3)x^2-(4-3z)x+4=0

$$(3y+3) { x }^{ 2 } -(4-3z)x+4=0$$

Solve for y

$$y=-\frac{z}{x}-1+\frac{4}{3x}-\frac{4}{3x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 3y+3 by x^{2}.

$$3yx^{2}+3x^{2}-\left(4-3z\right)x+4=0$$

Use the distributive property to multiply 4-3z by x.

$$3yx^{2}+3x^{2}-\left(4x-3zx\right)+4=0$$

To find the opposite of 4x-3zx, find the opposite of each term.

$$3yx^{2}+3x^{2}-4x+3zx+4=0$$

Subtract 3x^{2} from both sides. Anything subtracted from zero gives its negation.

$$3yx^{2}-4x+3zx+4=-3x^{2}$$

Add 4x to both sides.

$$3yx^{2}+3zx+4=-3x^{2}+4x$$

Subtract 3zx from both sides.

$$3yx^{2}+4=-3x^{2}+4x-3zx$$

Subtract 4 from both sides.

$$3yx^{2}=-3x^{2}+4x-3zx-4$$

The equation is in standard form.

$$3x^{2}y=-3x^{2}-3xz+4x-4$$

Divide both sides by 3x^{2}.

$$\frac{3x^{2}y}{3x^{2}}=\frac{-3x^{2}-3xz+4x-4}{3x^{2}}$$

Dividing by 3x^{2} undoes the multiplication by 3x^{2}.

$$y=\frac{-3x^{2}-3xz+4x-4}{3x^{2}}$$

Divide -3x^{2}+4x-3zx-4 by 3x^{2}.

$$y=\frac{-xz+\frac{4x}{3}-\frac{4}{3}}{x^{2}}-1$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{9z^{2}-48y-24z-32}-3z+4}{6\left(y+1\right)}\text{; }x=\frac{-\sqrt{9z^{2}-48y-24z-32}-3z+4}{6\left(y+1\right)}\text{, }&y\neq -1\\x=\frac{4}{4-3z}\text{, }&y=-1\text{ and }z\neq \frac{4}{3}\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{9z^{2}-48y-24z-32}-3z+4}{6\left(y+1\right)}\text{; }x=\frac{-\sqrt{9z^{2}-48y-24z-32}-3z+4}{6\left(y+1\right)}\text{, }&y\neq -1\text{ and }y\leq \frac{\left(4-3z\right)^{2}}{48}-1\\x=\frac{4}{4-3z}\text{, }&y=-1\text{ and }z\neq \frac{4}{3}\end{matrix}\right.$$