Solve 7/x^2+x+3/x^2-x=6/1-x^2

$$\frac { 7 } { x ^ { 2 } + x } + \frac { 3 } { x ^ { 2 } - x } = \frac { 6 } { 1 - x ^ { 2 } }$$

Solve for x

$$x=\frac{1}{4}=0.25$$

Steps for Solving Linear Equation

Variable x cannot be equal to any of the values -1,0,1 since division by zero is not defined. Multiply both sides of the equation by x\left(x-1\right)\left(x+1\right), the least common multiple of x^{2}+x,x^{2}-x,1-x^{2}.

$$\left(x-1\right)\times 7+\left(x+1\right)\times 3=-x\times 6$$

Use the distributive property to multiply x-1 by 7.

$$7x-7+\left(x+1\right)\times 3=-x\times 6$$

Use the distributive property to multiply x+1 by 3.

$$7x-7+3x+3=-x\times 6$$

Combine 7x and 3x to get 10x.

$$10x-7+3=-x\times 6$$

Add -7 and 3 to get -4.

$$10x-4=-x\times 6$$

Multiply -1 and 6 to get -6.

$$10x-4=-6x$$

Add 6x to both sides.

$$10x-4+6x=0$$

Combine 10x and 6x to get 16x.

$$16x-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$16x=4$$

Divide both sides by 16.

$$x=\frac{4}{16}$$

Reduce the fraction \frac{4}{16} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{4}$$