Graph 3/(2x)+1/x<=1/2

$\frac{3}{2 x} + \frac{1}{x} \leq \frac{1}{2}$

Step 1

Simplify $\frac{3}{2 x} + \frac{1}{x}$ .

To write $\frac{1}{x}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{3}{2 x} + \frac{1}{x} \cdot \frac{2}{2} \leq \frac{1}{2}$

Write each expression with a common denominator of $2 x$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{x}$ by $\frac{2}{2}$ .

$\frac{3}{2 x} + \frac{2}{x \cdot 2} \leq \frac{1}{2}$

Reorder the factors of $x \cdot 2$ .

$\frac{3}{2 x} + \frac{2}{2 x} \leq \frac{1}{2}$

Combine the numerators over the common denominator.

$\frac{3 + 2}{2 x} \leq \frac{1}{2}$

Add $3$ and $2$ .

$\frac{5}{2 x} \leq \frac{1}{2}$

Step 2

Multiply both sides by $2 x$ .

$\frac{5}{2 x} (2 x) = \frac{1}{2} (2 x)$

Step 3

Simplify.

Simplify the left side.

Simplify $\frac{5}{2 x} (2 x)$ .

Rewrite using the commutative property of multiplication.

$2 \frac{5}{2 x} x = \frac{1}{2} (2 x)$

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

$2 \frac{5}{2 (x)} x = \frac{1}{2} (2 x)$

Cancel the common factor.

$2 \frac{5}{2 x} x = \frac{1}{2} (2 x)$

Rewrite the expression.

$\frac{5}{x} x = \frac{1}{2} (2 x)$

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{5}{x} x = \frac{1}{2} (2 x)$

Rewrite the expression.

$5 = \frac{1}{2} (2 x)$

Simplify the right side.

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

$5 = \frac{1}{2} (2 (x))$

Cancel the common factor.

$5 = \frac{1}{2} (2 x)$

Rewrite the expression.

$5 = x$

Step 4

Rewrite the equation as $x = 5$ .

$x = 5$

Step 5

Find the domain of $\frac{5}{2 x} - \frac{1}{2}$ .

Set the denominator in $\frac{5}{2 x}$ equal to $0$ to find where the expression is undefined.

$2 x = 0$

Divide each term in $2 x = 0$ by $2$ and simplify.

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x = 0$

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.

$x < 0$

$0 < x < 5$

$x > 5$

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 $x < 0$ 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.

$\frac{3}{2 (- 2)} + \frac{1}{- 2} \leq \frac{1}{2}$

The left side $- 1.25$ is less than the right side $0.5$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 5$ and see if this value makes the original inequality true.

$x = 2$

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

$\frac{3}{2 (2)} + \frac{1}{2} \leq \frac{1}{2}$

The left side $1.25$ is greater than the right side $0.5$ , which means that the given statement is false.

False

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$\frac{3}{2 (8)} + \frac{1}{8} \leq \frac{1}{2}$

The left side $0.3125$ is less than the right side $0.5$ , which means that the given statement is always true.

True

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

$x < 0$ True

$0 < x < 5$ False

$x > 5$ True

$x < 0$ True

$0 < x < 5$ False

$x > 5$ True

Step 8

The solution consists of all of the true intervals.

$x < 0$ or $x \geq 5$

Step 9

Graph 3/(2x)+1/x<=1/2

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