Trigonometry Examples

$\frac{4}{3 x} + \frac{7}{x} < \frac{5}{9}$

Step 1

Simplify $\frac{4}{3 x} + \frac{7}{x}$ .

Tap for more steps...

Step 1.1

To write $\frac{7}{x}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{4}{3 x} + \frac{7}{x} \cdot \frac{3}{3} < \frac{5}{9}$

Step 1.2

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

Tap for more steps...

Step 1.2.1

Multiply $\frac{7}{x}$ by $\frac{3}{3}$ .

$\frac{4}{3 x} + \frac{7 \cdot 3}{x \cdot 3} < \frac{5}{9}$

Step 1.2.2

Reorder the factors of $x \cdot 3$ .

$\frac{4}{3 x} + \frac{7 \cdot 3}{3 x} < \frac{5}{9}$

Step 1.3

Combine the numerators over the common denominator.

$\frac{4 + 7 \cdot 3}{3 x} < \frac{5}{9}$

Step 1.4

Simplify the numerator.

Tap for more steps...

Step 1.4.1

Multiply $7$ by $3$ .

$\frac{4 + 21}{3 x} < \frac{5}{9}$

Step 1.4.2

Add $4$ and $21$ .

$\frac{25}{3 x} < \frac{5}{9}$

Step 2

Multiply both sides by $3 x$ .

$\frac{25}{3 x} (3 x) = \frac{5}{9} (3 x)$

Step 3

Simplify.

Tap for more steps...

Step 3.1

Simplify the left side.

Tap for more steps...

Step 3.1.1

Simplify $\frac{25}{3 x} (3 x)$ .

Tap for more steps...

Step 3.1.1.1

Rewrite using the commutative property of multiplication.

$3 \frac{25}{3 x} x = \frac{5}{9} (3 x)$

Step 3.1.1.2

Cancel the common factor of $3$ .

Tap for more steps...

Step 3.1.1.2.1

Factor $3$ out of $3 x$ .

$3 \frac{25}{3 (x)} x = \frac{5}{9} (3 x)$

Step 3.1.1.2.2

Cancel the common factor.

$3 \frac{25}{3 x} x = \frac{5}{9} (3 x)$

Step 3.1.1.2.3

Rewrite the expression.

$\frac{25}{x} x = \frac{5}{9} (3 x)$

Step 3.1.1.3

Cancel the common factor of $x$ .

Tap for more steps...

Step 3.1.1.3.1

Cancel the common factor.

$\frac{25}{x} x = \frac{5}{9} (3 x)$

Step 3.1.1.3.2

Rewrite the expression.

$25 = \frac{5}{9} (3 x)$

Step 3.2

Simplify the right side.

Tap for more steps...

Step 3.2.1

Simplify $\frac{5}{9} (3 x)$ .

Tap for more steps...

Step 3.2.1.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 3.2.1.1.1

Factor $3$ out of $9$ .

$25 = \frac{5}{3 (3)} (3 x)$

Step 3.2.1.1.2

Factor $3$ out of $3 x$ .

$25 = \frac{5}{3 (3)} (3 (x))$

Step 3.2.1.1.3

Cancel the common factor.

$25 = \frac{5}{3 \cdot 3} (3 x)$

Step 3.2.1.1.4

Rewrite the expression.

$25 = \frac{5}{3} x$

Step 3.2.1.2

Combine $\frac{5}{3}$ and $x$ .

$25 = \frac{5 x}{3}$

Step 4

Solve for $x$ .

Tap for more steps...

Step 4.1

Rewrite the equation as $\frac{5 x}{3} = 25$ .

$\frac{5 x}{3} = 25$

Step 4.2

Multiply both sides of the equation by $\frac{3}{5}$ .

$\frac{3}{5} \cdot \frac{5 x}{3} = \frac{3}{5} \cdot 25$

Step 4.3

Simplify both sides of the equation.

Tap for more steps...

Step 4.3.1

Simplify the left side.

Tap for more steps...

Step 4.3.1.1

Simplify $\frac{3}{5} \cdot \frac{5 x}{3}$ .

Tap for more steps...

Step 4.3.1.1.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 4.3.1.1.1.1

Cancel the common factor.

$\frac{3}{5} \cdot \frac{5 x}{3} = \frac{3}{5} \cdot 25$

Step 4.3.1.1.1.2

Rewrite the expression.

$\frac{1}{5} (5 x) = \frac{3}{5} \cdot 25$

Step 4.3.1.1.2

Cancel the common factor of $5$ .

Tap for more steps...

Step 4.3.1.1.2.1

Factor $5$ out of $5 x$ .

$\frac{1}{5} (5 (x)) = \frac{3}{5} \cdot 25$

Step 4.3.1.1.2.2

Cancel the common factor.

$\frac{1}{5} (5 x) = \frac{3}{5} \cdot 25$

Step 4.3.1.1.2.3

Rewrite the expression.

$x = \frac{3}{5} \cdot 25$

Step 4.3.2

Simplify the right side.

Tap for more steps...

Step 4.3.2.1

Simplify $\frac{3}{5} \cdot 25$ .

Tap for more steps...

Step 4.3.2.1.1

Cancel the common factor of $5$ .

Tap for more steps...

Step 4.3.2.1.1.1

Factor $5$ out of $25$ .

$x = \frac{3}{5} \cdot (5 (5))$

Step 4.3.2.1.1.2

Cancel the common factor.

$x = \frac{3}{5} \cdot (5 \cdot 5)$

Step 4.3.2.1.1.3

Rewrite the expression.

$x = 3 \cdot 5$

Step 4.3.2.1.2

Multiply $3$ by $5$ .

$x = 15$

Step 5

Find the domain of $\frac{25}{3 x} - \frac{5}{9}$ .

Tap for more steps...

Step 5.1

Set the denominator in $\frac{25}{3 x}$ equal to $0$ to find where the expression is undefined.

$3 x = 0$

Step 5.2

Divide each term in $3 x = 0$ by $3$ and simplify.

Tap for more steps...

Step 5.2.1

Divide each term in $3 x = 0$ by $3$ .

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

Step 5.2.2

Simplify the left side.

Tap for more steps...

Step 5.2.2.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 5.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.1.2

Divide $x$ by $1$ .

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

Step 5.2.3

Simplify the right side.

Tap for more steps...

Step 5.2.3.1

Divide $0$ by $3$ .

$x = 0$

Step 5.3

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

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

Step 6

Use each root to create test intervals.

$x < 0$

$0 < x < 15$

$x > 15$

Step 7

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

Tap for more steps...

Step 7.1

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

Tap for more steps...

Step 7.1.1

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

$x = - 2$

Step 7.1.2

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

$\frac{4}{3 (- 2)} + \frac{7}{- 2} < \frac{5}{9}$

Step 7.1.3

The left side $- 4.1\overline{6}$ is less than the right side $0.\overline{5}$ , which means that the given statement is always true.

True

Step 7.2

Test a value on the interval $0 < x < 15$ to see if it makes the inequality true.

Tap for more steps...

Step 7.2.1

Choose a value on the interval $0 < x < 15$ and see if this value makes the original inequality true.

$x = 2$

Step 7.2.2

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

$\frac{4}{3 (2)} + \frac{7}{2} < \frac{5}{9}$

Step 7.2.3

The left side $4.1\overline{6}$ is not less than the right side $0.\overline{5}$ , which means that the given statement is false.

False

Step 7.3

Test a value on the interval $x > 15$ to see if it makes the inequality true.

Tap for more steps...

Step 7.3.1

Choose a value on the interval $x > 15$ and see if this value makes the original inequality true.

$x = 18$

Step 7.3.2

Replace $x$ with $18$ in the original inequality.

$\frac{4}{3 (18)} + \frac{7}{18} < \frac{5}{9}$

Step 7.3.3

The left side $0.4\overline{629}$ is less than the right side $0.\overline{5}$ , which means that the given statement is always true.

True

Step 7.4

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

$x < 0$ True

$0 < x < 15$ False

$x > 15$ True

$x < 0$ True

$0 < x < 15$ False

$x > 15$ True

Step 8

The solution consists of all of the true intervals.

$x < 0$ or $x > 15$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$x < 0 or x > 15$

Interval Notation:

$(- \infty, 0) \cup (15, \infty)$

Step 10