Algebra Examples

$\frac{12}{x^{2} + 2 x} < \frac{3}{x^{2} + 4 x + 4}$

Step 1

Multiply both sides by $x^{2} + 2 x$ .

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

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Cancel the common factor of $x^{2} + 2 x$ .

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Step 2.1.1.1

Cancel the common factor.

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

Step 2.1.1.2

Rewrite the expression.

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

Step 2.2

Simplify the right side.

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Step 2.2.1

Simplify $\frac{3}{x^{2} + 4 x + 4} (x^{2} + 2 x)$ .

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Step 2.2.1.1

Factor using the perfect square rule.

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Step 2.2.1.1.1

Rewrite $4$ as $2^{2}$ .

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

Step 2.2.1.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Step 2.2.1.1.3

Rewrite the polynomial.

$12 = \frac{3}{x^{2} + 2 \cdot x \cdot 2 + 2^{2}} (x^{2} + 2 x)$

Step 2.2.1.1.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 2$ .

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

Step 2.2.1.2

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

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

Step 2.2.1.3

Factor $x$ out of $x^{2} + 2 x$ .

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Step 2.2.1.3.1

Factor $x$ out of $x^{2}$ .

$12 = \frac{3 (x \cdot x + 2 x)}{{(x + 2)}^{2}}$

Step 2.2.1.3.2

Factor $x$ out of $2 x$ .

$12 = \frac{3 (x \cdot x + x \cdot 2)}{{(x + 2)}^{2}}$

Step 2.2.1.3.3

Factor $x$ out of $x \cdot x + x \cdot 2$ .

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

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

Step 2.2.1.4

Cancel the common factor of $x + 2$ and ${(x + 2)}^{2}$ .

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Step 2.2.1.4.1

Factor $x + 2$ out of $3 x (x + 2)$ .

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

Step 2.2.1.4.2

Cancel the common factors.

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Step 2.2.1.4.2.1

Factor $x + 2$ out of ${(x + 2)}^{2}$ .

$12 = \frac{(x + 2) (3 x)}{(x + 2) (x + 2)}$

Step 2.2.1.4.2.2

Cancel the common factor.

$12 = \frac{(x + 2) (3 x)}{(x + 2) (x + 2)}$

Step 2.2.1.4.2.3

Rewrite the expression.

$12 = \frac{3 x}{x + 2}$

Step 3

Solve for $x$ .

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Step 3.1

Rewrite the equation as $\frac{3 x}{x + 2} = 12$ .

$\frac{3 x}{x + 2} = 12$

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$x + 2, 1$

Step 3.2.2

Remove parentheses.

$x + 2, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$x + 2$

Step 3.3

Multiply each term in $\frac{3 x}{x + 2} = 12$ by $x + 2$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{3 x}{x + 2} = 12$ by $x + 2$ .

$\frac{3 x}{x + 2} (x + 2) = 12 (x + 2)$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $x + 2$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{3 x}{x + 2} (x + 2) = 12 (x + 2)$

Step 3.3.2.1.2

Rewrite the expression.

$3 x = 12 (x + 2)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

$3 x = 12 x + 12 \cdot 2$

Step 3.3.3.2

Multiply $12$ by $2$ .

$3 x = 12 x + 24$

Step 3.4

Solve the equation.

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Step 3.4.1

Move all terms containing $x$ to the left side of the equation.

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Step 3.4.1.1

Subtract $12 x$ from both sides of the equation.

$3 x - 12 x = 24$

Step 3.4.1.2

Subtract $12 x$ from $3 x$ .

$- 9 x = 24$

Step 3.4.2

Divide each term in $- 9 x = 24$ by $- 9$ and simplify.

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Step 3.4.2.1

Divide each term in $- 9 x = 24$ by $- 9$ .

$\frac{- 9 x}{- 9} = \frac{24}{- 9}$

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Cancel the common factor of $- 9$ .

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Step 3.4.2.2.1.1

Cancel the common factor.

$\frac{- 9 x}{- 9} = \frac{24}{- 9}$

Step 3.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{24}{- 9}$

Step 3.4.2.3

Simplify the right side.

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Step 3.4.2.3.1

Cancel the common factor of $24$ and $- 9$ .

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Step 3.4.2.3.1.1

Factor $3$ out of $24$ .

$x = \frac{3 (8)}{- 9}$

Step 3.4.2.3.1.2

Cancel the common factors.

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Step 3.4.2.3.1.2.1

Factor $3$ out of $- 9$ .

$x = \frac{3 \cdot 8}{3 \cdot - 3}$

Step 3.4.2.3.1.2.2

Cancel the common factor.

$x = \frac{3 \cdot 8}{3 \cdot - 3}$

Step 3.4.2.3.1.2.3

Rewrite the expression.

$x = \frac{8}{- 3}$

Step 3.4.2.3.2

Move the negative in front of the fraction.

$x = - \frac{8}{3}$

Step 4

Find the domain of $\frac{12}{x (x + 2)} - \frac{3}{{(x + 2)}^{2}}$ .

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Step 4.1

Set the denominator in $\frac{12}{x (x + 2)}$ equal to $0$ to find where the expression is undefined.

$x (x + 2) = 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x + 2 = 0$

Step 4.2.2

Set $x$ equal to $0$ .

$x = 0$

Step 4.2.3

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 4.2.3.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 4.2.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4.2.4

The final solution is all the values that make $x (x + 2) = 0$ true.

$x = 0, - 2$

Step 4.3

Set the denominator in $\frac{3}{{(x + 2)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x + 2)}^{2} = 0$

Step 4.4

Solve for $x$ .

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Step 4.4.1

Set the $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 4.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 4.5

The domain is all values of $x$ that make the expression defined.

$(- \infty, - 2) \cup (- 2, 0) \cup (0, \infty)$

Step 5

Use each root to create test intervals.

$x < - \frac{8}{3}$

$- \frac{8}{3} < x < - 2$

$- 2 < x < 0$

$x > 0$

Step 6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 6.1

Test a value on the interval $x < - \frac{8}{3}$ to see if it makes the inequality true.

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Step 6.1.1

Choose a value on the interval $x < - \frac{8}{3}$ and see if this value makes the original inequality true.

$x = - 5$

Step 6.1.2

Replace $x$ with $- 5$ in the original inequality.

$\frac{12}{{(- 5)}^{2} + 2 (- 5)} < \frac{3}{{(- 5)}^{2} + 4 (- 5) + 4}$

Step 6.1.3

The left side $0.8$ is not less than the right side $0.\overline{3}$ , which means that the given statement is false.

False

Step 6.2

Test a value on the interval $- \frac{8}{3} < x < - 2$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $- \frac{8}{3} < x < - 2$ and see if this value makes the original inequality true.

$x = - 2.33$

Step 6.2.2

Replace $x$ with $- 2.33$ in the original inequality.

$\frac{12}{{(- 2.33)}^{2} + 2 (- 2.33)} < \frac{3}{{(- 2.33)}^{2} + 4 (- 2.33) + 4}$

Step 6.2.3

The left side $15.60671088$ is less than the right side $27.54820936$ , which means that the given statement is always true.

True

Step 6.3

Test a value on the interval $- 2 < x < 0$ to see if it makes the inequality true.

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Step 6.3.1

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Step 6.3.2

Replace $x$ with $- 1$ in the original inequality.

$\frac{12}{{(- 1)}^{2} + 2 (- 1)} < \frac{3}{{(- 1)}^{2} + 4 (- 1) + 4}$

Step 6.3.3

The left side $- 12$ is less than the right side $3$ , which means that the given statement is always true.

True

Step 6.4

Test a value on the interval $x > 0$ to see if it makes the inequality true.

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Step 6.4.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 6.4.2

Replace $x$ with $2$ in the original inequality.

$\frac{12}{{(2)}^{2} + 2 (2)} < \frac{3}{{(2)}^{2} + 4 (2) + 4}$

Step 6.4.3

The left side $1.5$ is not less than the right side $0.1875$ , which means that the given statement is false.

False

Step 6.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < - \frac{8}{3}$ False

$- \frac{8}{3} < x < - 2$ True

$- 2 < x < 0$ True

$x > 0$ False

$x < - \frac{8}{3}$ False

$- \frac{8}{3} < x < - 2$ True

$- 2 < x < 0$ True

$x > 0$ False

Step 7

The solution consists of all of the true intervals.

$- \frac{8}{3} < x < - 2$ or $- 2 < x < 0$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$- \frac{8}{3} < x < - 2 or - 2 < x < 0$

Interval Notation:

$(- \frac{8}{3}, - 2) \cup (- 2, 0)$

Step 9