Solve 18x^3+24x^2-5xy-4=8

$$18 x ^ { 3 } + 24 x ^ { 2 } - 5 x y - 4 = 8$$

Solve for y

$$y=\frac{18x^{2}}{5}+\frac{24x}{5}-\frac{12}{5x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 18x^{3} from both sides.

$$24x^{2}-5xy-4=8-18x^{3}$$

Subtract 24x^{2} from both sides.

$$-5xy-4=8-18x^{3}-24x^{2}$$

Add 4 to both sides.

$$-5xy=8-18x^{3}-24x^{2}+4$$

Add 8 and 4 to get 12.

$$-5xy=12-18x^{3}-24x^{2}$$

The equation is in standard form.

$$\left(-5x\right)y=12-24x^{2}-18x^{3}$$

Divide both sides by -5x.

$$\frac{\left(-5x\right)y}{-5x}=\frac{12-24x^{2}-18x^{3}}{-5x}$$

Dividing by -5x undoes the multiplication by -5x.

$$y=\frac{12-24x^{2}-18x^{3}}{-5x}$$

Divide 12-18x^{3}-24x^{2} by -5x.

$$y=\frac{18x^{2}}{5}+\frac{24x}{5}-\frac{12}{5x}$$