Finite Math Examples

$\frac{14}{x^{2} - 3 x} - \frac{8}{x} > - \frac{10}{x - 3}$

Step 1

Add $\frac{10}{x - 3}$ to both sides of the inequality.

$\frac{14}{x^{2} - 3 x} - \frac{8}{x} + \frac{10}{x - 3} > 0$

Step 2

Simplify $\frac{14}{x^{2} - 3 x} - \frac{8}{x} + \frac{10}{x - 3}$ .

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Step 2.1

Factor $x$ out of $x^{2} - 3 x$ .

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Step 2.1.1

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

$\frac{14}{x \cdot x - 3 x} - \frac{8}{x} + \frac{10}{x - 3} > 0$

Step 2.1.2

Factor $x$ out of $- 3 x$ .

$\frac{14}{x \cdot x + x \cdot - 3} - \frac{8}{x} + \frac{10}{x - 3} > 0$

Step 2.1.3

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

$\frac{14}{x (x - 3)} - \frac{8}{x} + \frac{10}{x - 3} > 0$

Step 2.2

To write $\frac{10}{x - 3}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{14}{x (x - 3)} + \frac{10}{x - 3} \cdot \frac{x}{x} - \frac{8}{x} > 0$

Step 2.3

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

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Step 2.3.1

Multiply $\frac{10}{x - 3}$ by $\frac{x}{x}$ .

$\frac{14}{x (x - 3)} + \frac{10 x}{(x - 3) x} - \frac{8}{x} > 0$

Step 2.3.2

Reorder the factors of $(x - 3) x$ .

$\frac{14}{x (x - 3)} + \frac{10 x}{x (x - 3)} - \frac{8}{x} > 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{14 + 10 x}{x (x - 3)} - \frac{8}{x} > 0$

Step 2.5

Factor $2$ out of $14 + 10 x$ .

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Step 2.5.1

Factor $2$ out of $14$ .

$\frac{2 (7) + 10 x}{x (x - 3)} - \frac{8}{x} > 0$

Step 2.5.2

Factor $2$ out of $10 x$ .

$\frac{2 (7) + 2 (5 x)}{x (x - 3)} - \frac{8}{x} > 0$

Step 2.5.3

Factor $2$ out of $2 (7) + 2 (5 x)$ .

$\frac{2 (7 + 5 x)}{x (x - 3)} - \frac{8}{x} > 0$

Step 2.6

To write $- \frac{8}{x}$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

$\frac{2 (7 + 5 x)}{x (x - 3)} - \frac{8}{x} \cdot \frac{x - 3}{x - 3} > 0$

Step 2.7

Multiply $\frac{8}{x}$ by $\frac{x - 3}{x - 3}$ .

$\frac{2 (7 + 5 x)}{x (x - 3)} - \frac{8 (x - 3)}{x (x - 3)} > 0$

Step 2.8

Combine the numerators over the common denominator.

$\frac{2 (7 + 5 x) - 8 (x - 3)}{x (x - 3)} > 0$

Step 2.9

Simplify the numerator.

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Step 2.9.1

Factor $2$ out of $2 (7 + 5 x) - 8 (x - 3)$ .

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Step 2.9.1.1

Factor $2$ out of $- 8 (x - 3)$ .

$\frac{2 (7 + 5 x) + 2 (- 4 (x - 3))}{x (x - 3)} > 0$

Step 2.9.1.2

Factor $2$ out of $2 (7 + 5 x) + 2 (- 4 (x - 3))$ .

$\frac{2 (7 + 5 x - 4 (x - 3))}{x (x - 3)} > 0$

Step 2.9.2

Apply the distributive property.

$\frac{2 (7 + 5 x - 4 x - 4 \cdot - 3)}{x (x - 3)} > 0$

Step 2.9.3

Multiply $- 4$ by $- 3$ .

$\frac{2 (7 + 5 x - 4 x + 12)}{x (x - 3)} > 0$

Step 2.9.4

Add $7$ and $12$ .

$\frac{2 (5 x - 4 x + 19)}{x (x - 3)} > 0$

Step 2.9.5

Subtract $4 x$ from $5 x$ .

$\frac{2 (x + 19)}{x (x - 3)} > 0$

Step 3

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

$x = 0$

$x + 19 = 0$

$x - 3 = 0$

Step 4

Subtract $19$ from both sides of the equation.

$x = - 19$

Step 5

Add $3$ to both sides of the equation.

$x = 3$

Step 6

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

$x = 0$

$x = - 19$

$x = 3$

Step 7

Consolidate the solutions.

$x = 0, - 19, 3$

Step 8

Find the domain of $\frac{2 (x + 19)}{x (x - 3)}$ .

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Step 8.1

Set the denominator in $\frac{2 (x + 19)}{x (x - 3)}$ equal to $0$ to find where the expression is undefined.

$x (x - 3) = 0$

Step 8.2

Solve for $x$ .

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Step 8.2.1

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

$x = 0$

$x - 3 = 0$

Step 8.2.2

Set $x$ equal to $0$ .

$x = 0$

Step 8.2.3

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 8.2.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 8.2.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 8.2.4

The final solution is all the values that make $x (x - 3) = 0$ true.

$x = 0, 3$

Step 8.3

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

$(- \infty, 0) \cup (0, 3) \cup (3, \infty)$

Step 9

Use each root to create test intervals.

$x < - 19$

$- 19 < x < 0$

$0 < x < 3$

$x > 3$

Step 10

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 10.1

Test a value on the interval $x < - 19$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < - 19$ and see if this value makes the original inequality true.

$x = - 22$

Step 10.1.2

Replace $x$ with $- 22$ in the original inequality.

$\frac{14}{{(- 22)}^{2} - 3 \cdot - 22} - \frac{8}{- 22} > - \frac{10}{(- 22) - 3}$

Step 10.1.3

The left side $0.38\overline{90}$ is not greater than the right side $0.4$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $- 19 < x < 0$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 19 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 10.2.2

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

$\frac{14}{{(- 2)}^{2} - 3 \cdot - 2} - \frac{8}{- 2} > - \frac{10}{(- 2) - 3}$

Step 10.2.3

The left side $5.4$ is greater than the right side $2$ , which means that the given statement is always true.

True

Step 10.3

Test a value on the interval $0 < x < 3$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $0 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Step 10.3.2

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

$\frac{14}{{(2)}^{2} - 3 \cdot 2} - \frac{8}{2} > - \frac{10}{(2) - 3}$

Step 10.3.3

The left side $- 11$ is not greater than the right side $10$ , which means that the given statement is false.

False

Step 10.4

Test a value on the interval $x > 3$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Step 10.4.2

Replace $x$ with $6$ in the original inequality.

$\frac{14}{{(6)}^{2} - 3 \cdot 6} - \frac{8}{6} > - \frac{10}{(6) - 3}$

Step 10.4.3

The left side $- 0.\overline{5}$ is greater than the right side $- 3.\overline{3}$ , which means that the given statement is always true.

True

Step 10.5

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

$x < - 19$ False

$- 19 < x < 0$ True

$0 < x < 3$ False

$x > 3$ True

$x < - 19$ False

$- 19 < x < 0$ True

$0 < x < 3$ False

$x > 3$ True

Step 11

The solution consists of all of the true intervals.

$- 19 < x < 0$ or $x > 3$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$- 19 < x < 0 or x > 3$

Interval Notation:

$(- 19, 0) \cup (3, \infty)$

Step 13