Solve 1+4/x-y/x^2=0

$$1 + \frac { 4 } { x } - \frac { y } { x ^ { 2 } } = 0$$

Solve for y

$$y=x\left(x+4\right)$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x^{2}, the least common multiple of x,x^{2}.

$$x^{2}+x\times 4-y=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

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

Subtract x\times 4 from both sides.

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

Multiply -1 and 4 to get -4.

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

Divide both sides by -1.

$$\frac{-y}{-1}=-\frac{x\left(x+4\right)}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$y=-\frac{x\left(x+4\right)}{-1}$$

Divide -x\left(4+x\right) by -1.

$$y=x\left(x+4\right)$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=-\sqrt{y+4}-2\text{, }&\text{unconditionally}\\x=\sqrt{y+4}-2\text{, }&y\neq 0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=-\sqrt{y+4}-2\text{, }&y\geq -4\\x=\sqrt{y+4}-2\text{, }&y\neq 0\text{ and }y\geq -4\end{matrix}\right.$$