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

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

Solve for x

$$x=-2$$

Steps Using Factoring

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

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

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

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

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

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

To find the opposite of -x-x^{2}, find the opposite of each term.

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

Combine 3x and x to get 4x.

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

Subtract x from both sides.

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

Combine 4x and -x to get 3x.

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

Add 2 to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor x^{2}+3x+2 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

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

To find equation solutions, solve x+1=0 and x+2=0.

$$x=-1 x=-2$$

Variable x cannot be equal to -1.

$$x=-2$$

Steps Using Factoring By Grouping

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

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

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

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

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

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

To find the opposite of -x-x^{2}, find the opposite of each term.

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

Combine 3x and x to get 4x.

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

Subtract x from both sides.

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

Combine 4x and -x to get 3x.

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

Add 2 to both sides.

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

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

Rewrite x^{2}+3x+2 as \left(x^{2}+x\right)+\left(2x+2\right).

$$\left(x^{2}+x\right)+\left(2x+2\right)$$

Factor out x in the first and 2 in the second group.

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

Factor out common term x+1 by using distributive property.

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

To find equation solutions, solve x+1=0 and x+2=0.

$$x=-1 x=-2$$

Variable x cannot be equal to -1.

$$x=-2$$

Steps Using the Quadratic Formula

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

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

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

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

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

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

To find the opposite of -x-x^{2}, find the opposite of each term.

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

Combine 3x and x to get 4x.

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

Subtract x from both sides.

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

Combine 4x and -x to get 3x.

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

Add 2 to both sides.

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

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.

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

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 2}}{2}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 2}}{2}$$

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8}}{2}$$

Add 9 to -8.

$$x=\frac{-3±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-3±1}{2}$$

Now solve the equation x=\frac{-3±1}{2} when ± is plus. Add -3 to 1.

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

Divide -2 by 2.

$$x=-1$$

Now solve the equation x=\frac{-3±1}{2} when ± is minus. Subtract 1 from -3.

$$x=-\frac{4}{2}$$

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=-1 x=-2$$

Variable x cannot be equal to -1.

$$x=-2$$

Steps for Completing the Square

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

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

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

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

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

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

To find the opposite of -x-x^{2}, find the opposite of each term.

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

Combine 3x and x to get 4x.

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

Subtract x from both sides.

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

Combine 4x and -x to get 3x.

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

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.

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

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.

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

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

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

Add -2 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{1}{4}$$

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

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

Subtract \frac{3}{2} from both sides of the equation.

$$x=-1 x=-2$$

Variable x cannot be equal to -1.

$$x=-2$$