Solve 2/x-2x/x+1=2/x+1

$$\frac { 2 } { x } - \frac { 2 x } { x + 1 } = \frac { 2 } { x + 1 }$$

Solve for x

$$x=1$$

Steps by Finding Square Root

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

$$\left(x+1\right)\times 2-x\times 2x=x\times 2$$

Use the distributive property to multiply x+1 by 2.

$$2x+2-x\times 2x=x\times 2$$

Multiply x and x to get x^{2}.

$$2x+2-x^{2}\times 2=x\times 2$$

Subtract x\times 2 from both sides.

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

Combine 2x and -x\times 2 to get 0.

$$2-x^{2}\times 2=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

$$x^{2}\times 2=\frac{-2}{-1}$$

Fraction \frac{-2}{-1} can be simplified to 2 by removing the negative sign from both the numerator and the denominator.

$$x^{2}\times 2=2$$

Divide both sides by 2.

$$x^{2}=\frac{2}{2}$$

Divide 2 by 2 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Variable x cannot be equal to -1.

$$x=1$$

Steps Using the Quadratic Formula

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

$$\left(x+1\right)\times 2-x\times 2x=x\times 2$$

Use the distributive property to multiply x+1 by 2.

$$2x+2-x\times 2x=x\times 2$$

Multiply x and x to get x^{2}.

$$2x+2-x^{2}\times 2=x\times 2$$

Subtract x\times 2 from both sides.

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

Combine 2x and -x\times 2 to get 0.

$$2-x^{2}\times 2=0$$

Multiply -1 and 2 to get -2.

$$2-2x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$-2x^{2}+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 0 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8\times 2}}{2\left(-2\right)}$$

Multiply 8 times 2.

$$x=\frac{0±\sqrt{16}}{2\left(-2\right)}$$

Take the square root of 16.

$$x=\frac{0±4}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{0±4}{-4}$$

Now solve the equation x=\frac{0±4}{-4} when ± is plus. Divide 4 by -4.

$$x=-1$$

Now solve the equation x=\frac{0±4}{-4} when ± is minus. Divide -4 by -4.

$$x=1$$

The equation is now solved.

$$x=-1 x=1$$

Variable x cannot be equal to -1.

$$x=1$$