Algebra Examples

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

Step 1

Subtract $\frac{4}{x + 2}$ from both sides of the equation.

$\frac{- 2 x}{x + 2} + \frac{x}{3} - \frac{4}{x + 2} = 0$

Step 2

Simplify $\frac{- 2 x}{x + 2} + \frac{x}{3} - \frac{4}{x + 2}$ .

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Step 2.1

Move the negative in front of the fraction.

$- \frac{2 x}{x + 2} + \frac{x}{3} - \frac{4}{x + 2} = 0$

Step 2.2

To write $- \frac{2 x}{x + 2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- \frac{2 x}{x + 2} \cdot \frac{3}{3} + \frac{x}{3} - \frac{4}{x + 2} = 0$

Step 2.3

To write $\frac{x}{3}$ as a fraction with a common denominator, multiply by $\frac{x + 2}{x + 2}$ .

$- \frac{2 x}{x + 2} \cdot \frac{3}{3} + \frac{x}{3} \cdot \frac{x + 2}{x + 2} - \frac{4}{x + 2} = 0$

Step 2.4

Write each expression with a common denominator of $(x + 2) \cdot 3$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.4.1

Multiply $\frac{2 x}{x + 2}$ by $\frac{3}{3}$ .

$- \frac{2 x \cdot 3}{(x + 2) \cdot 3} + \frac{x}{3} \cdot \frac{x + 2}{x + 2} - \frac{4}{x + 2} = 0$

Step 2.4.2

Multiply $\frac{x}{3}$ by $\frac{x + 2}{x + 2}$ .

$- \frac{2 x \cdot 3}{(x + 2) \cdot 3} + \frac{x (x + 2)}{3 (x + 2)} - \frac{4}{x + 2} = 0$

Step 2.4.3

Reorder the factors of $(x + 2) \cdot 3$ .

$- \frac{2 x \cdot 3}{3 (x + 2)} + \frac{x (x + 2)}{3 (x + 2)} - \frac{4}{x + 2} = 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{- 2 x \cdot 3 + x (x + 2)}{3 (x + 2)} - \frac{4}{x + 2} = 0$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Factor $x$ out of $- 2 x \cdot 3 + x (x + 2)$ .

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Step 2.6.1.1

Factor $x$ out of $- 2 x \cdot 3$ .

$\frac{x (- 2 \cdot 3) + x (x + 2)}{3 (x + 2)} - \frac{4}{x + 2} = 0$

Step 2.6.1.2

Factor $x$ out of $x (- 2 \cdot 3) + x (x + 2)$ .

$\frac{x (- 2 \cdot 3 + x + 2)}{3 (x + 2)} - \frac{4}{x + 2} = 0$

Step 2.6.2

Multiply $- 2$ by $3$ .

$\frac{x (- 6 + x + 2)}{3 (x + 2)} - \frac{4}{x + 2} = 0$

Step 2.6.3

Add $- 6$ and $2$ .

$\frac{x (x - 4)}{3 (x + 2)} - \frac{4}{x + 2} = 0$

Step 2.7

To write $- \frac{4}{x + 2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{x (x - 4)}{3 (x + 2)} - \frac{4}{x + 2} \cdot \frac{3}{3} = 0$

Step 2.8

Write each expression with a common denominator of $3 (x + 2)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.8.1

Multiply $\frac{4}{x + 2}$ by $\frac{3}{3}$ .

$\frac{x (x - 4)}{3 (x + 2)} - \frac{4 \cdot 3}{(x + 2) \cdot 3} = 0$

Step 2.8.2

Reorder the factors of $(x + 2) \cdot 3$ .

$\frac{x (x - 4)}{3 (x + 2)} - \frac{4 \cdot 3}{3 (x + 2)} = 0$

Step 2.9

Combine the numerators over the common denominator.

$\frac{x (x - 4) - 4 \cdot 3}{3 (x + 2)} = 0$

Step 2.10

Simplify the numerator.

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Step 2.10.1

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 4 - 4 \cdot 3}{3 (x + 2)} = 0$

Step 2.10.2

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot - 4 - 4 \cdot 3}{3 (x + 2)} = 0$

Step 2.10.3

Move $- 4$ to the left of $x$ .

$\frac{x^{2} - 4 \cdot x - 4 \cdot 3}{3 (x + 2)} = 0$

Step 2.10.4

Multiply $- 4$ by $3$ .

$\frac{x^{2} - 4 x - 12}{3 (x + 2)} = 0$

Step 2.10.5

Factor $x^{2} - 4 x - 12$ using the AC method.

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Step 2.10.5.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 12$ and whose sum is $- 4$ .

$- 6, 2$

Step 2.10.5.2

Write the factored form using these integers.

$\frac{(x - 6) (x + 2)}{3 (x + 2)} = 0$

Step 2.11

Cancel the common factor of $x + 2$ .

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Step 2.11.1

Cancel the common factor.

$\frac{(x - 6) (x + 2)}{3 (x + 2)} = 0$

Step 2.11.2

Rewrite the expression.

$\frac{x - 6}{3} = 0$

Step 3

Set the numerator equal to zero.

$x - 6 = 0$

Step 4

Add $6$ to both sides of the equation.

$x = 6$