Solve z^2y-xz^3-1=0

$$z ^ { 2 } y - x z ^ { 3 } - 1 = 0$$

Solve for x

$$x=\frac{yz^{2}-1}{z^{3}}$$

$$z\neq 0$$

Steps for Solving Linear Equation

Add 1 to both sides. Anything plus zero gives itself.

$$z^{2}y-xz^{3}=1$$

Subtract z^{2}y from both sides.

$$-xz^{3}=1-z^{2}y$$

The equation is in standard form.

$$\left(-z^{3}\right)x=1-yz^{2}$$

Divide both sides by -z^{3}.

$$\frac{\left(-z^{3}\right)x}{-z^{3}}=\frac{1-yz^{2}}{-z^{3}}$$

Dividing by -z^{3} undoes the multiplication by -z^{3}.

$$x=\frac{1-yz^{2}}{-z^{3}}$$

Divide 1-z^{2}y by -z^{3}.

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

Solve for y

$$y=xz+\frac{1}{z^{2}}$$

$$z\neq 0$$

Steps for Solving Linear Equation

Add 1 to both sides. Anything plus zero gives itself.

$$z^{2}y-xz^{3}=1$$

Add xz^{3} to both sides.

$$z^{2}y=1+xz^{3}$$

The equation is in standard form.

$$z^{2}y=xz^{3}+1$$

Divide both sides by z^{2}.

$$\frac{z^{2}y}{z^{2}}=\frac{xz^{3}+1}{z^{2}}$$

Dividing by z^{2} undoes the multiplication by z^{2}.

$$y=\frac{xz^{3}+1}{z^{2}}$$

Divide 1+xz^{3} by z^{2}.

$$y=xz+\frac{1}{z^{2}}$$