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

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

Solve for x

$$x=3$$

$$x=0$$

Steps Using the Quadratic Formula

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

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

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

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

Combine 2x and x to get 3x.

$$3x-2+1=\left(x-1\right)\left(x+1\right)$$

Add -2 and 1 to get -1.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$3x-1=x^{2}-1$$

Subtract x^{2} from both sides.

$$3x-1-x^{2}=-1$$

Add 1 to both sides.

$$3x-1-x^{2}+1=0$$

Add -1 and 1 to get 0.

$$3x-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+3x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}}}{2\left(-1\right)}$$

Take the square root of 3^{2}.

$$x=\frac{-3±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-3±3}{-2}$$

Now solve the equation x=\frac{-3±3}{-2} when ± is plus. Add -3 to 3.

$$x=\frac{0}{-2}$$

Divide 0 by -2.

$$x=0$$

Now solve the equation x=\frac{-3±3}{-2} when ± is minus. Subtract 3 from -3.

$$x=-\frac{6}{-2}$$

Divide -6 by -2.

$$x=3$$

The equation is now solved.

$$x=0 x=3$$

Steps for Completing the Square

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

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

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

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

Combine 2x and x to get 3x.

$$3x-2+1=\left(x-1\right)\left(x+1\right)$$

Add -2 and 1 to get -1.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$3x-1=x^{2}-1$$

Subtract x^{2} from both sides.

$$3x-1-x^{2}=-1$$

Add 1 to both sides.

$$3x-x^{2}=-1+1$$

Add -1 and 1 to get 0.

$$3x-x^{2}=0$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+3x=0$$

Divide both sides by -1.

$$\frac{-x^{2}+3x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{3}{-1}x=\frac{0}{-1}$$

Divide 3 by -1.

$$x^{2}-3x=\frac{0}{-1}$$

Divide 0 by -1.

$$x^{2}-3x=0$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=\frac{9}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$x-\frac{3}{2}=\frac{3}{2} x-\frac{3}{2}=-\frac{3}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$x=3 x=0$$