Solve x/x+1-3x+3/x-2=0

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

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=-\frac{1}{2}=-0.5$$

Steps Using Factoring By Grouping

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+1,x.

$$xx-\left(x+1\right)\left(3x+3\right)+x\left(x+1\right)\left(-2\right)=0$$

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

$$x^{2}-\left(x+1\right)\left(3x+3\right)+x\left(x+1\right)\left(-2\right)=0$$

Use the distributive property to multiply x+1 by 3x+3 and combine like terms.

$$x^{2}-\left(3x^{2}+6x+3\right)+x\left(x+1\right)\left(-2\right)=0$$

To find the opposite of 3x^{2}+6x+3, find the opposite of each term.

$$x^{2}-3x^{2}-6x-3+x\left(x+1\right)\left(-2\right)=0$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-6x-3+x\left(x+1\right)\left(-2\right)=0$$

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

$$-2x^{2}-6x-3+\left(x^{2}+x\right)\left(-2\right)=0$$

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

$$-2x^{2}-6x-3-2x^{2}-2x=0$$

Combine -2x^{2} and -2x^{2} to get -4x^{2}.

$$-4x^{2}-6x-3-2x=0$$

Combine -6x and -2x to get -8x.

$$-4x^{2}-8x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-4\left(-3\right)=12$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 12.

$$-1,-12 -2,-6 -3,-4$$

Calculate the sum for each pair.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

The solution is the pair that gives sum -8.

$$a=-2 b=-6$$

Rewrite -4x^{2}-8x-3 as \left(-4x^{2}-2x\right)+\left(-6x-3\right).

$$\left(-4x^{2}-2x\right)+\left(-6x-3\right)$$

Factor out 2x in the first and 3 in the second group.

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

Factor out common term -2x-1 by using distributive property.

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

To find equation solutions, solve -2x-1=0 and 2x+3=0.

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

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+1,x.

$$xx-\left(x+1\right)\left(3x+3\right)+x\left(x+1\right)\left(-2\right)=0$$

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

$$x^{2}-\left(x+1\right)\left(3x+3\right)+x\left(x+1\right)\left(-2\right)=0$$

Use the distributive property to multiply x+1 by 3x+3 and combine like terms.

$$x^{2}-\left(3x^{2}+6x+3\right)+x\left(x+1\right)\left(-2\right)=0$$

To find the opposite of 3x^{2}+6x+3, find the opposite of each term.

$$x^{2}-3x^{2}-6x-3+x\left(x+1\right)\left(-2\right)=0$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-6x-3+x\left(x+1\right)\left(-2\right)=0$$

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

$$-2x^{2}-6x-3+\left(x^{2}+x\right)\left(-2\right)=0$$

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

$$-2x^{2}-6x-3-2x^{2}-2x=0$$

Combine -2x^{2} and -2x^{2} to get -4x^{2}.

$$-4x^{2}-6x-3-2x=0$$

Combine -6x and -2x to get -8x.

$$-4x^{2}-8x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -8 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-4\right)\left(-3\right)}}{2\left(-4\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-4\right)\left(-3\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-8\right)±\sqrt{64+16\left(-3\right)}}{2\left(-4\right)}$$

Multiply 16 times -3.

$$x=\frac{-\left(-8\right)±\sqrt{64-48}}{2\left(-4\right)}$$

Add 64 to -48.

$$x=\frac{-\left(-8\right)±\sqrt{16}}{2\left(-4\right)}$$

Take the square root of 16.

$$x=\frac{-\left(-8\right)±4}{2\left(-4\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±4}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{8±4}{-8}$$

Now solve the equation x=\frac{8±4}{-8} when ± is plus. Add 8 to 4.

$$x=\frac{12}{-8}$$

Reduce the fraction \frac{12}{-8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{8±4}{-8} when ± is minus. Subtract 4 from 8.

$$x=\frac{4}{-8}$$

Reduce the fraction \frac{4}{-8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

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+1,x.

$$xx-\left(x+1\right)\left(3x+3\right)+x\left(x+1\right)\left(-2\right)=0$$

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

$$x^{2}-\left(x+1\right)\left(3x+3\right)+x\left(x+1\right)\left(-2\right)=0$$

Use the distributive property to multiply x+1 by 3x+3 and combine like terms.

$$x^{2}-\left(3x^{2}+6x+3\right)+x\left(x+1\right)\left(-2\right)=0$$

To find the opposite of 3x^{2}+6x+3, find the opposite of each term.

$$x^{2}-3x^{2}-6x-3+x\left(x+1\right)\left(-2\right)=0$$

Combine x^{2} and -3x^{2} to get -2x^{2}.

$$-2x^{2}-6x-3+x\left(x+1\right)\left(-2\right)=0$$

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

$$-2x^{2}-6x-3+\left(x^{2}+x\right)\left(-2\right)=0$$

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

$$-2x^{2}-6x-3-2x^{2}-2x=0$$

Combine -2x^{2} and -2x^{2} to get -4x^{2}.

$$-4x^{2}-6x-3-2x=0$$

Combine -6x and -2x to get -8x.

$$-4x^{2}-8x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$-4x^{2}-8x=3$$

Divide both sides by -4.

$$\frac{-4x^{2}-8x}{-4}=\frac{3}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{8}{-4}\right)x=\frac{3}{-4}$$

Divide -8 by -4.

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

Divide 3 by -4.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 1.

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

Add -\frac{3}{4} to 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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