Solve xy=x^2+1/x^2-4-1

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

Solve for y

$$y=\frac{5}{x\left(x^{2}-4\right)}$$

$$x\neq 0\text{ and }|x|\neq 2$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(x-2\right)\left(x+2\right).

$$xy\left(x-2\right)\left(x+2\right)=x^{2}+1+\left(x-2\right)\left(x+2\right)\left(-1\right)$$

Use the distributive property to multiply xy by x-2.

$$\left(yx^{2}-2xy\right)\left(x+2\right)=x^{2}+1+\left(x-2\right)\left(x+2\right)\left(-1\right)$$

Use the distributive property to multiply yx^{2}-2xy by x+2 and combine like terms.

$$yx^{3}-4xy=x^{2}+1+\left(x-2\right)\left(x+2\right)\left(-1\right)$$

Use the distributive property to multiply x-2 by x+2 and combine like terms.

$$yx^{3}-4xy=x^{2}+1+\left(x^{2}-4\right)\left(-1\right)$$

Use the distributive property to multiply x^{2}-4 by -1.

$$yx^{3}-4xy=x^{2}+1-x^{2}+4$$

Combine x^{2} and -x^{2} to get 0.

$$yx^{3}-4xy=1+4$$

Add 1 and 4 to get 5.

$$yx^{3}-4xy=5$$

Combine all terms containing y.

$$\left(x^{3}-4x\right)y=5$$

Divide both sides by x^{3}-4x.

$$\frac{\left(x^{3}-4x\right)y}{x^{3}-4x}=\frac{5}{x^{3}-4x}$$

Dividing by x^{3}-4x undoes the multiplication by x^{3}-4x.

$$y=\frac{5}{x^{3}-4x}$$

Divide 5 by x^{3}-4x.

$$y=\frac{5}{x\left(x^{2}-4\right)}$$