Solve y^4=-3x+16/2

$$y ^ { 4 } = \frac { - 3 x + 16 } { 2 }$$

Solve for x

$$x=\frac{16-2y^{4}}{3}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$2y^{4}=-3x+16$$

Swap sides so that all variable terms are on the left hand side.

$$-3x+16=2y^{4}$$

Subtract 16 from both sides.

$$-3x=2y^{4}-16$$

Divide both sides by -3.

$$\frac{-3x}{-3}=\frac{2y^{4}-16}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x=\frac{2y^{4}-16}{-3}$$

Divide 2y^{4}-16 by -3.

$$x=\frac{16-2y^{4}}{3}$$

Solve for y (complex solution)

$$y=\frac{2^{\frac{3}{4}}i\sqrt[4]{16-3x}}{2}$$

$$y=\frac{2^{\frac{3}{4}}\sqrt[4]{16-3x}}{2}$$

$$y=-\frac{2^{\frac{3}{4}}\sqrt[4]{16-3x}}{2}$$

$$y=-\frac{2^{\frac{3}{4}}i\sqrt[4]{16-3x}}{2}$$

Solve for y

$$y=\frac{2^{\frac{3}{4}}\sqrt[4]{16-3x}}{2}$$

$$y=-\frac{2^{\frac{3}{4}}\sqrt[4]{16-3x}}{2}\text{, }x\leq \frac{16}{3}$$