Solve x^3+5x^2y+xy-5=0

$$x ^ { 3 } + 5 x ^ { 2 } y + x y - 5 = 0$$

Solve for y

$$y=-\frac{x^{3}-5}{x\left(5x+1\right)}$$

$$x\neq -\frac{1}{5}\text{ and }x\neq 0$$

Steps for Solving Linear Equation

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

$$5x^{2}y+xy-5=-x^{3}$$

Add 5 to both sides.

$$5x^{2}y+xy=-x^{3}+5$$

Combine all terms containing y.

$$\left(5x^{2}+x\right)y=-x^{3}+5$$

The equation is in standard form.

$$\left(5x^{2}+x\right)y=5-x^{3}$$

Divide both sides by 5x^{2}+x.

$$\frac{\left(5x^{2}+x\right)y}{5x^{2}+x}=\frac{5-x^{3}}{5x^{2}+x}$$

Dividing by 5x^{2}+x undoes the multiplication by 5x^{2}+x.

$$y=\frac{5-x^{3}}{5x^{2}+x}$$

Divide -x^{3}+5 by 5x^{2}+x.

$$y=\frac{5-x^{3}}{x\left(5x+1\right)}$$