Solve for x 1/x+1/3<1/5

$\frac{1}{x} + \frac{1}{3} < \frac{1}{5}$

Step 1

Move all terms not containing $x$ to the right side of the inequality.

Subtract $\frac{1}{3}$ from both sides of the inequality.

$\frac{1}{x} < \frac{1}{5} - \frac{1}{3}$

To write $\frac{1}{5}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{x} < \frac{1}{5} \cdot \frac{3}{3} - \frac{1}{3}$

To write $- \frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{1}{x} < \frac{1}{5} \cdot \frac{3}{3} - \frac{1}{3} \cdot \frac{5}{5}$

Write each expression with a common denominator of $15$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{5}$ by $\frac{3}{3}$ .

$\frac{1}{x} < \frac{3}{5 \cdot 3} - \frac{1}{3} \cdot \frac{5}{5}$

Multiply $5$ by $3$ .

$\frac{1}{x} < \frac{3}{15} - \frac{1}{3} \cdot \frac{5}{5}$

Multiply $\frac{1}{3}$ by $\frac{5}{5}$ .

$\frac{1}{x} < \frac{3}{15} - \frac{5}{3 \cdot 5}$

Multiply $3$ by $5$ .

$\frac{1}{x} < \frac{3}{15} - \frac{5}{15}$

Combine the numerators over the common denominator.

$\frac{1}{x} < \frac{3 - 5}{15}$

Subtract $5$ from $3$ .

$\frac{1}{x} < \frac{- 2}{15}$

Move the negative in front of the fraction.

$\frac{1}{x} < - \frac{2}{15}$

Step 2

Multiply both sides by $x$ .

$\frac{1}{x} x = - \frac{2}{15} x$

Step 3

Simplify.

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{1}{x} x = - \frac{2}{15} x$

Rewrite the expression.

$1 = - \frac{2}{15} x$

Simplify the right side.

Simplify $- \frac{2}{15} x$ .

Combine $x$ and $\frac{2}{15}$ .

$1 = - \frac{x \cdot 2}{15}$

Move $2$ to the left of $x$ .

$1 = - \frac{2 x}{15}$

Step 4

Solve for $x$ .

Rewrite the equation as $- \frac{2 x}{15} = 1$ .

$- \frac{2 x}{15} = 1$

Multiply both sides of the equation by $- \frac{15}{2}$ .

$- \frac{15}{2} (- \frac{2 x}{15}) = - \frac{15}{2} \cdot 1$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{15}{2} (- \frac{2 x}{15})$ .

Cancel the common factor of $15$ .

Move the leading negative in $- \frac{15}{2}$ into the numerator.

$\frac{- 15}{2} (- \frac{2 x}{15}) = - \frac{15}{2} \cdot 1$

Move the leading negative in $- \frac{2 x}{15}$ into the numerator.

$\frac{- 15}{2} \cdot \frac{- 2 x}{15} = - \frac{15}{2} \cdot 1$

Factor $15$ out of $- 15$ .

$\frac{15 (- 1)}{2} \cdot \frac{- 2 x}{15} = - \frac{15}{2} \cdot 1$

Cancel the common factor.

$\frac{15 \cdot - 1}{2} \cdot \frac{- 2 x}{15} = - \frac{15}{2} \cdot 1$

Rewrite the expression.

$\frac{- 1}{2} (- 2 x) = - \frac{15}{2} \cdot 1$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2 x$ .

$\frac{- 1}{2} (2 (- x)) = - \frac{15}{2} \cdot 1$

Cancel the common factor.

$\frac{- 1}{2} (2 (- x)) = - \frac{15}{2} \cdot 1$

Rewrite the expression.

$- - x = - \frac{15}{2} \cdot 1$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 x = - \frac{15}{2} \cdot 1$

Multiply $x$ by $1$ .

$x = - \frac{15}{2} \cdot 1$

Simplify the right side.

Multiply $- 1$ by $1$ .

$x = - \frac{15}{2}$

Step 5

Find the domain of $\frac{1}{x} + \frac{2}{15}$ .

Set the denominator in $\frac{1}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

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.

$x < - \frac{15}{2}$

$- \frac{15}{2} < x < 0$

$x > 0$

Step 7

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

Test a value on the interval $x < - \frac{15}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{15}{2}$ and see if this value makes the original inequality true.

$x = - 10$

Replace $x$ with $- 10$ in the original inequality.

$\frac{1}{- 10} + \frac{1}{3} < \frac{1}{5}$

The left side $0.2\overline{3}$ is not less than the right side $0.2$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{15}{2} < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{15}{2} < x < 0$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$\frac{1}{- 4} + \frac{1}{3} < \frac{1}{5}$

The left side $0.08\overline{3}$ is less than the right side $0.2$ , which means that the given statement is always true.

True

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

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

$x = 2$

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

$\frac{1}{2} + \frac{1}{3} < \frac{1}{5}$

The left side $0.8\overline{3}$ is not less than the right side $0.2$ , which means that the given statement is false.

False

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

$x < - \frac{15}{2}$ False

$- \frac{15}{2} < x < 0$ True

$x > 0$ False

$x < - \frac{15}{2}$ False

$- \frac{15}{2} < x < 0$ True

$x > 0$ False

Step 8

The solution consists of all of the true intervals.

$- \frac{15}{2} < x < 0$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$- \frac{15}{2} < x < 0$

Interval Notation:

$(- \frac{15}{2}, 0)$

Step 10

Solve for x 1/x+1/3<1/5

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