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

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

Solve for x

$$x=-4$$

$$x=1$$

Steps Using Factoring By Grouping

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by 3\left(x+2\right), the least common multiple of x+2,3.

$$3\times 2+3\left(x+2\right)\left(-\frac{1}{3}\right)=\left(x+2\right)x$$

Multiply 3 and 2 to get 6.

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

Multiply 3 and -\frac{1}{3} to get -1.

$$6-\left(x+2\right)=\left(x+2\right)x$$

To find the opposite of x+2, find the opposite of each term.

$$6-x-2=\left(x+2\right)x$$

Subtract 2 from 6 to get 4.

$$4-x=\left(x+2\right)x$$

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

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

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

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

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

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

$$a+b=-3 ab=-4=-4$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=1 b=-4$$

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

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

Factor out x in the first and 4 in the second group.

$$x\left(-x+1\right)+4\left(-x+1\right)$$

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

$$\left(-x+1\right)\left(x+4\right)$$

To find equation solutions, solve -x+1=0 and x+4=0.

$$x=1 x=-4$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by 3\left(x+2\right), the least common multiple of x+2,3.

$$3\times 2+3\left(x+2\right)\left(-\frac{1}{3}\right)=\left(x+2\right)x$$

Multiply 3 and 2 to get 6.

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

Multiply 3 and -\frac{1}{3} to get -1.

$$6-\left(x+2\right)=\left(x+2\right)x$$

To find the opposite of x+2, find the opposite of each term.

$$6-x-2=\left(x+2\right)x$$

Subtract 2 from 6 to get 4.

$$4-x=\left(x+2\right)x$$

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

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

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

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.

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

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

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 4}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 4}}{2\left(-1\right)}$$

Multiply 4 times 4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\left(-1\right)}$$

Add 9 to 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$x=\frac{-\left(-3\right)±5}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±5}{-2}$$

Now solve the equation x=\frac{3±5}{-2} when ± is plus. Add 3 to 5.

$$x=\frac{8}{-2}$$

Divide 8 by -2.

$$x=-4$$

Now solve the equation x=\frac{3±5}{-2} when ± is minus. Subtract 5 from 3.

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-4 x=1$$

Steps for Completing the Square

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by 3\left(x+2\right), the least common multiple of x+2,3.

$$3\times 2+3\left(x+2\right)\left(-\frac{1}{3}\right)=\left(x+2\right)x$$

Multiply 3 and 2 to get 6.

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

Multiply 3 and -\frac{1}{3} to get -1.

$$6-\left(x+2\right)=\left(x+2\right)x$$

To find the opposite of x+2, find the opposite of each term.

$$6-x-2=\left(x+2\right)x$$

Subtract 2 from 6 to get 4.

$$4-x=\left(x+2\right)x$$

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

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

Subtract x^{2} from both sides.

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

Subtract 2x from both sides.

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

Combine -x and -2x to get -3x.

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

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -3 by -1.

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

Divide -4 by -1.

$$x^{2}+3x=4$$

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.

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

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

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

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

$$x^{2}+3x+\frac{9}{4}=\frac{25}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{5}{2} x+\frac{3}{2}=-\frac{5}{2}$$

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

$$x=1 x=-4$$