Solve y^-3quad1/2x+13=9

$$y ^ { - 3 } \quad \frac { 1 } { 2 } x + 13 = 9$$

Solve for x

$$x=-8y^{3}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Subtract 13 from both sides.

$$y^{-3}\times \frac{1}{2}x=9-13$$

Subtract 13 from 9 to get -4.

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

The equation is in standard form.

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

Divide both sides by \frac{1}{2}y^{-3}.

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

Dividing by \frac{1}{2}y^{-3} undoes the multiplication by \frac{1}{2}y^{-3}.

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

Divide -4 by \frac{1}{2}y^{-3}.

$$x=-8y^{3}$$

Solve for y (complex solution)

$$y=-\frac{\sqrt[3]{x}}{2}$$

$$y=\frac{e^{\frac{\pi i}{3}}\sqrt[3]{x}}{2}$$

$$y=\frac{e^{\frac{5\pi i}{3}}\sqrt[3]{x}}{2}\text{, }x\neq 0$$

Solve for y

$$y=-\frac{\sqrt[3]{x}}{2}$$

$$x\neq 0$$