Algebra Examples

$\frac{1}{x} + \frac{1}{x - 10} \geq \frac{2}{24}$

Step 1

Subtract $\frac{2}{24}$ from both sides of the inequality.

$\frac{1}{x} + \frac{1}{x - 10} - \frac{2}{24} \geq 0$

Step 2

Simplify $\frac{1}{x} + \frac{1}{x - 10} - \frac{2}{24}$ .

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Step 2.1

Find the common denominator.

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Step 2.1.1

Multiply $\frac{1}{x}$ by $\frac{(x - 10) \cdot 24}{(x - 10) \cdot 24}$ .

$\frac{1}{x} \cdot \frac{(x - 10) \cdot 24}{(x - 10) \cdot 24} + \frac{1}{x - 10} - \frac{2}{24} \geq 0$

Step 2.1.2

Multiply $\frac{1}{x}$ by $\frac{(x - 10) \cdot 24}{(x - 10) \cdot 24}$ .

$\frac{(x - 10) \cdot 24}{x ((x - 10) \cdot 24)} + \frac{1}{x - 10} - \frac{2}{24} \geq 0$

Step 2.1.3

Multiply $\frac{1}{x - 10}$ by $\frac{x \cdot 24}{x \cdot 24}$ .

$\frac{(x - 10) \cdot 24}{x ((x - 10) \cdot 24)} + \frac{1}{x - 10} \cdot \frac{x \cdot 24}{x \cdot 24} - \frac{2}{24} \geq 0$

Step 2.1.4

Multiply $\frac{1}{x - 10}$ by $\frac{x \cdot 24}{x \cdot 24}$ .

$\frac{(x - 10) \cdot 24}{x ((x - 10) \cdot 24)} + \frac{x \cdot 24}{(x - 10) (x \cdot 24)} - \frac{2}{24} \geq 0$

Step 2.1.5

Multiply $\frac{2}{24}$ by $\frac{x (x - 10)}{x (x - 10)}$ .

$\frac{(x - 10) \cdot 24}{x ((x - 10) \cdot 24)} + \frac{x \cdot 24}{(x - 10) (x \cdot 24)} - (\frac{2}{24} \cdot \frac{x (x - 10)}{x (x - 10)}) \geq 0$

Step 2.1.6

Multiply $\frac{2}{24}$ by $\frac{x (x - 10)}{x (x - 10)}$ .

$\frac{(x - 10) \cdot 24}{x ((x - 10) \cdot 24)} + \frac{x \cdot 24}{(x - 10) (x \cdot 24)} - \frac{2 (x (x - 10))}{24 (x (x - 10))} \geq 0$

Step 2.1.7

Reorder the factors of $x ((x - 10) \cdot 24)$ .

$\frac{(x - 10) \cdot 24}{24 x (x - 10)} + \frac{x \cdot 24}{(x - 10) (x \cdot 24)} - \frac{2 (x (x - 10))}{24 (x (x - 10))} \geq 0$

Step 2.1.8

Reorder the factors of $(x - 10) (x \cdot 24)$ .

$\frac{(x - 10) \cdot 24}{24 x (x - 10)} + \frac{x \cdot 24}{24 x (x - 10)} - \frac{2 (x (x - 10))}{24 (x (x - 10))} \geq 0$

Step 2.1.9

Reorder the factors of $24 (x (x - 10))$ .

$\frac{(x - 10) \cdot 24}{24 x (x - 10)} + \frac{x \cdot 24}{24 x (x - 10)} - \frac{2 (x (x - 10))}{24 x (x - 10)} \geq 0$

Step 2.2

Combine the numerators over the common denominator.

$\frac{(x - 10) \cdot 24 + x \cdot 24 - 2 (x (x - 10))}{24 x (x - 10)} \geq 0$

Step 2.3

Simplify each term.

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Step 2.3.1

Apply the distributive property.

$\frac{x \cdot 24 - 10 \cdot 24 + x \cdot 24 - 2 (x (x - 10))}{24 x (x - 10)} \geq 0$

Step 2.3.2

Move $24$ to the left of $x$ .

$\frac{24 \cdot x - 10 \cdot 24 + x \cdot 24 - 2 (x (x - 10))}{24 x (x - 10)} \geq 0$

Step 2.3.3

Multiply $- 10$ by $24$ .

$\frac{24 \cdot x - 240 + x \cdot 24 - 2 (x (x - 10))}{24 x (x - 10)} \geq 0$

Step 2.3.4

Move $24$ to the left of $x$ .

$\frac{24 x - 240 + 24 \cdot x - 2 (x (x - 10))}{24 x (x - 10)} \geq 0$

Step 2.3.5

Apply the distributive property.

$\frac{24 x - 240 + 24 x - 2 (x \cdot x + x \cdot - 10)}{24 x (x - 10)} \geq 0$

Step 2.3.6

Multiply $x$ by $x$ .

$\frac{24 x - 240 + 24 x - 2 (x^{2} + x \cdot - 10)}{24 x (x - 10)} \geq 0$

Step 2.3.7

Move $- 10$ to the left of $x$ .

$\frac{24 x - 240 + 24 x - 2 (x^{2} - 10 \cdot x)}{24 x (x - 10)} \geq 0$

Step 2.3.8

Apply the distributive property.

$\frac{24 x - 240 + 24 x - 2 x^{2} - 2 (- 10 x)}{24 x (x - 10)} \geq 0$

Step 2.3.9

Multiply $- 10$ by $- 2$ .

$\frac{24 x - 240 + 24 x - 2 x^{2} + 20 x}{24 x (x - 10)} \geq 0$

Step 2.4

Add $24 x$ and $24 x$ .

$\frac{48 x - 240 - 2 x^{2} + 20 x}{24 x (x - 10)} \geq 0$

Step 2.5

Add $48 x$ and $20 x$ .

$\frac{68 x - 240 - 2 x^{2}}{24 x (x - 10)} \geq 0$

Step 2.6

Cancel the common factor of $68 x - 240 - 2 x^{2}$ and $24$ .

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Step 2.6.1

Factor $2$ out of $68 x$ .

$\frac{2 (34 x) - 240 - 2 x^{2}}{24 x (x - 10)} \geq 0$

Step 2.6.2

Factor $2$ out of $- 240$ .

$\frac{2 (34 x) + 2 (- 120) - 2 x^{2}}{24 x (x - 10)} \geq 0$

Step 2.6.3

Factor $2$ out of $2 (34 x) + 2 (- 120)$ .

$\frac{2 (34 x - 120) - 2 x^{2}}{24 x (x - 10)} \geq 0$

Step 2.6.4

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

$\frac{2 (34 x - 120) + 2 (- x^{2})}{24 x (x - 10)} \geq 0$

Step 2.6.5

Factor $2$ out of $2 (34 x - 120) + 2 (- x^{2})$ .

$\frac{2 (34 x - 120 - x^{2})}{24 x (x - 10)} \geq 0$

Step 2.6.6

Cancel the common factors.

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Step 2.6.6.1

Factor $2$ out of $24 x (x - 10)$ .

$\frac{2 (34 x - 120 - x^{2})}{2 (12 x (x - 10))} \geq 0$

Step 2.6.6.2

Cancel the common factor.

$\frac{2 (34 x - 120 - x^{2})}{2 (12 x (x - 10))} \geq 0$

Step 2.6.6.3

Rewrite the expression.

$\frac{34 x - 120 - x^{2}}{12 x (x - 10)} \geq 0$

Step 2.7

Factor by grouping.

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Step 2.7.1

Reorder terms.

$\frac{- x^{2} + 34 x - 120}{12 x (x - 10)} \geq 0$

Step 2.7.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 120 = 120$ and whose sum is $b = 34$ .

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Step 2.7.2.1

Factor $34$ out of $34 x$ .

$\frac{- x^{2} + 34 (x) - 120}{12 x (x - 10)} \geq 0$

Step 2.7.2.2

Rewrite $34$ as $4$ plus $30$

$\frac{- x^{2} + (4 + 30) x - 120}{12 x (x - 10)} \geq 0$

Step 2.7.2.3

Apply the distributive property.

$\frac{- x^{2} + 4 x + 30 x - 120}{12 x (x - 10)} \geq 0$

Step 2.7.3

Factor out the greatest common factor from each group.

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Step 2.7.3.1

Group the first two terms and the last two terms.

$\frac{(- x^{2} + 4 x) + 30 x - 120}{12 x (x - 10)} \geq 0$

Step 2.7.3.2

Factor out the greatest common factor (GCF) from each group.

$\frac{x (- x + 4) - 30 (- x + 4)}{12 x (x - 10)} \geq 0$

Step 2.7.4

Factor the polynomial by factoring out the greatest common factor, $- x + 4$ .

$\frac{(- x + 4) (x - 30)}{12 x (x - 10)} \geq 0$

Step 2.8

Factor $- 1$ out of $- x$ .

$\frac{(- (x) + 4) (x - 30)}{12 x (x - 10)} \geq 0$

Step 2.9

Rewrite $4$ as $- 1 (- 4)$ .

$\frac{(- (x) - 1 (- 4)) (x - 30)}{12 x (x - 10)} \geq 0$

Step 2.10

Factor $- 1$ out of $- (x) - 1 (- 4)$ .

$\frac{- (x - 4) (x - 30)}{12 x (x - 10)} \geq 0$

Step 2.11

Rewrite $- (x - 4)$ as $- 1 (x - 4)$ .

$\frac{- 1 (x - 4) (x - 30)}{12 x (x - 10)} \geq 0$

Step 2.12

Move the negative in front of the fraction.

$- \frac{(x - 4) (x - 30)}{12 x (x - 10)} \geq 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$12 x = 0$

$x - 4 = 0$

$x - 30 = 0$

$x - 10 = 0$

Step 4

Divide each term in $12 x = 0$ by $12$ and simplify.

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Step 4.1

Divide each term in $12 x = 0$ by $12$ .

$\frac{12 x}{12} = \frac{0}{12}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $12$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{12 x}{12} = \frac{0}{12}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{12}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $12$ .

$x = 0$

Step 5

Add $4$ to both sides of the equation.

$x = 4$

Step 6

Add $30$ to both sides of the equation.

$x = 30$

Step 7

Add $10$ to both sides of the equation.

$x = 10$

Step 8

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

$x = 4$

$x = 30$

$x = 10$

Step 9

Consolidate the solutions.

$x = 0, 4, 30, 10$

Step 10

Find the domain of $- \frac{(x - 4) (x - 30)}{12 x (x - 10)}$ .

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Step 10.1

Set the denominator in $\frac{(x - 4) (x - 30)}{12 x (x - 10)}$ equal to $0$ to find where the expression is undefined.

$12 x (x - 10) = 0$

Step 10.2

Solve for $x$ .

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Step 10.2.1

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

$x = 0$

$x - 10 = 0$

Step 10.2.2

Set $x$ equal to $0$ .

$x = 0$

Step 10.2.3

Set $x - 10$ equal to $0$ and solve for $x$ .

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Step 10.2.3.1

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Step 10.2.3.2

Add $10$ to both sides of the equation.

$x = 10$

Step 10.2.4

The final solution is all the values that make $12 x (x - 10) = 0$ true.

$x = 0, 10$

Step 10.3

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

$(- \infty, 0) \cup (0, 10) \cup (10, \infty)$

Step 11

Use each root to create test intervals.

$x < 0$

$0 < x < 4$

$4 < x < 10$

$10 < x < 30$

$x > 30$

Step 12

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 12.1

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

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Step 12.1.1

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

$x = - 2$

Step 12.1.2

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

$\frac{1}{- 2} + \frac{1}{(- 2) - 10} \geq \frac{2}{24}$

Step 12.1.3

The left side $- 0.58\overline{3}$ is less than the right side $0.08\overline{3}$ , which means that the given statement is false.

False

Step 12.2

Test a value on the interval $0 < x < 4$ to see if it makes the inequality true.

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Step 12.2.1

Choose a value on the interval $0 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 12.2.2

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

$\frac{1}{2} + \frac{1}{(2) - 10} \geq \frac{2}{24}$

Step 12.2.3

The left side $0.375$ is greater than the right side $0.08\overline{3}$ , which means that the given statement is always true.

True

Step 12.3

Test a value on the interval $4 < x < 10$ to see if it makes the inequality true.

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Step 12.3.1

Choose a value on the interval $4 < x < 10$ and see if this value makes the original inequality true.

$x = 7$

Step 12.3.2

Replace $x$ with $7$ in the original inequality.

$\frac{1}{7} + \frac{1}{(7) - 10} \geq \frac{2}{24}$

Step 12.3.3

The left side $- 0.\overline{190476}$ is less than the right side $0.08\overline{3}$ , which means that the given statement is false.

False

Step 12.4

Test a value on the interval $10 < x < 30$ to see if it makes the inequality true.

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Step 12.4.1

Choose a value on the interval $10 < x < 30$ and see if this value makes the original inequality true.

$x = 12$

Step 12.4.2

Replace $x$ with $12$ in the original inequality.

$\frac{1}{12} + \frac{1}{(12) - 10} \geq \frac{2}{24}$

Step 12.4.3

The left side $0.58\overline{3}$ is greater than the right side $0.08\overline{3}$ , which means that the given statement is always true.

True

Step 12.5

Test a value on the interval $x > 30$ to see if it makes the inequality true.

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Step 12.5.1

Choose a value on the interval $x > 30$ and see if this value makes the original inequality true.

$x = 32$

Step 12.5.2

Replace $x$ with $32$ in the original inequality.

$\frac{1}{32} + \frac{1}{(32) - 10} \geq \frac{2}{24}$

Step 12.5.3

The left side $0.07670\overline{45}$ is less than the right side $0.08\overline{3}$ , which means that the given statement is false.

False

Step 12.6

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

$x < 0$ False

$0 < x < 4$ True

$4 < x < 10$ False

$10 < x < 30$ True

$x > 30$ False

$x < 0$ False

$0 < x < 4$ True

$4 < x < 10$ False

$10 < x < 30$ True

$x > 30$ False

Step 13

The solution consists of all of the true intervals.

$0 < x \leq 4$ or $10 < x \leq 30$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$0 < x \leq 4 or 10 < x \leq 30$

Interval Notation:

$(0, 4] \cup (10, 30]$

Step 15