Solve y^3-16y/2x+18=4-y/y^2+9y

$$\frac { y ^ { 3 } - 16 y } { 2 x + 18 } = \frac { 4 - y } { y ^ { 2 } + 9 y }$$

Solve for x

$$\left\{\begin{matrix}x=\frac{-\left(y+4\right)\left(y+9\right)y^{2}-18}{2}\text{, }&y\neq 0\text{ and }y\neq -9\text{ and }y\neq -4\\x\neq -9\text{, }&y=4\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable x cannot be equal to -9 since division by zero is not defined. Multiply both sides of the equation by 2y\left(x+9\right)\left(y+9\right), the least common multiple of 2x+18,y^{2}+9y.

$$\left(y+9\right)y\left(y^{3}-16y\right)=\left(2x+18\right)\left(4-y\right)$$

Use the distributive property to multiply y+9 by y.

$$\left(y^{2}+9y\right)\left(y^{3}-16y\right)=\left(2x+18\right)\left(4-y\right)$$

Use the distributive property to multiply y^{2}+9y by y^{3}-16y.

$$y^{5}-16y^{3}+9y^{4}-144y^{2}=\left(2x+18\right)\left(4-y\right)$$

Use the distributive property to multiply 2x+18 by 4-y.

$$y^{5}-16y^{3}+9y^{4}-144y^{2}=8x-2yx+72-18y$$

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

$$8x-2yx+72-18y=y^{5}-16y^{3}+9y^{4}-144y^{2}$$

Subtract 72 from both sides.

$$8x-2yx-18y=y^{5}-16y^{3}+9y^{4}-144y^{2}-72$$

Add 18y to both sides.

$$8x-2yx=y^{5}-16y^{3}+9y^{4}-144y^{2}-72+18y$$

Combine all terms containing x.

$$\left(8-2y\right)x=y^{5}-16y^{3}+9y^{4}-144y^{2}-72+18y$$

The equation is in standard form.

$$\left(8-2y\right)x=y^{5}+9y^{4}-16y^{3}-144y^{2}+18y-72$$

Divide both sides by 8-2y.

$$\frac{\left(8-2y\right)x}{8-2y}=\frac{\left(y-4\right)\left(y^{4}+13y^{3}+36y^{2}+18\right)}{8-2y}$$

Dividing by 8-2y undoes the multiplication by 8-2y.

$$x=\frac{\left(y-4\right)\left(y^{4}+13y^{3}+36y^{2}+18\right)}{8-2y}$$

Divide \left(-4+y\right)\left(18+36y^{2}+13y^{3}+y^{4}\right) by 8-2y.

$$x=-\frac{y^{4}}{2}-\frac{13y^{3}}{2}-18y^{2}-9$$

Variable x cannot be equal to -9.

$$x=-\frac{y^{4}}{2}-\frac{13y^{3}}{2}-18y^{2}-9\text{, }x\neq -9$$