Solve for p 8/(2p)+11/(4p)<1/2

$\frac{8}{2 p} + \frac{11}{4 p} < \frac{1}{2}$

Step 1

Simplify $\frac{8}{2 p} + \frac{11}{4 p}$ .

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8$ .

$\frac{2 \cdot 4}{2 p} + \frac{11}{4 p} < \frac{1}{2}$

Cancel the common factors.

Factor $2$ out of $2 p$ .

$\frac{2 \cdot 4}{2 (p)} + \frac{11}{4 p} < \frac{1}{2}$

Cancel the common factor.

$\frac{2 \cdot 4}{2 p} + \frac{11}{4 p} < \frac{1}{2}$

Rewrite the expression.

$\frac{4}{p} + \frac{11}{4 p} < \frac{1}{2}$

To write $\frac{4}{p}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{4}{p} \cdot \frac{4}{4} + \frac{11}{4 p} < \frac{1}{2}$

Write each expression with a common denominator of $p \cdot 4$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4}{p}$ by $\frac{4}{4}$ .

$\frac{4 \cdot 4}{p \cdot 4} + \frac{11}{4 p} < \frac{1}{2}$

Move $4$ to the left of $p$ .

$\frac{4 \cdot 4}{4 p} + \frac{11}{4 p} < \frac{1}{2}$

Combine the numerators over the common denominator.

$\frac{4 \cdot 4 + 11}{4 p} < \frac{1}{2}$

Simplify the numerator.

Multiply $4$ by $4$ .

$\frac{16 + 11}{4 p} < \frac{1}{2}$

Add $16$ and $11$ .

$\frac{27}{4 p} < \frac{1}{2}$

Step 2

Multiply both sides by $4 p$ .

$\frac{27}{4 p} (4 p) = \frac{1}{2} (4 p)$

Step 3

Simplify.

Simplify the left side.

Simplify $\frac{27}{4 p} (4 p)$ .

Rewrite using the commutative property of multiplication.

$4 \frac{27}{4 p} p = \frac{1}{2} (4 p)$

Cancel the common factor of $4$ .

Factor $4$ out of $4 p$ .

$4 \frac{27}{4 (p)} p = \frac{1}{2} (4 p)$

Cancel the common factor.

$4 \frac{27}{4 p} p = \frac{1}{2} (4 p)$

Rewrite the expression.

$\frac{27}{p} p = \frac{1}{2} (4 p)$

Cancel the common factor of $p$ .

Cancel the common factor.

$\frac{27}{p} p = \frac{1}{2} (4 p)$

Rewrite the expression.

$27 = \frac{1}{2} (4 p)$

Simplify the right side.

Cancel the common factor of $2$ .

Factor $2$ out of $4 p$ .

$27 = \frac{1}{2} (2 (2 p))$

Cancel the common factor.

$27 = \frac{1}{2} (2 (2 p))$

Rewrite the expression.

$27 = 2 p$

Step 4

Solve for $p$ .

Rewrite the equation as $2 p = 27$ .

$2 p = 27$

Divide each term in $2 p = 27$ by $2$ and simplify.

Divide each term in $2 p = 27$ by $2$ .

$\frac{2 p}{2} = \frac{27}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 p}{2} = \frac{27}{2}$

Divide $p$ by $1$ .

$p = \frac{27}{2}$

Step 5

Find the domain of $\frac{27}{4 p} - \frac{1}{2}$ .

Set the denominator in $\frac{27}{4 p}$ equal to $0$ to find where the expression is undefined.

$4 p = 0$

Divide each term in $4 p = 0$ by $4$ and simplify.

Divide each term in $4 p = 0$ by $4$ .

$\frac{4 p}{4} = \frac{0}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 p}{4} = \frac{0}{4}$

Divide $p$ by $1$ .

$p = \frac{0}{4}$

Simplify the right side.

Divide $0$ by $4$ .

$p = 0$

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

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

Step 6

Use each root to create test intervals.

$p < 0$

$0 < p < \frac{27}{2}$

$p > \frac{27}{2}$

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 $p < 0$ to see if it makes the inequality true.

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

$p = - 2$

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

$\frac{8}{2 (- 2)} + \frac{11}{4 (- 2)} < \frac{1}{2}$

The left side $- 3.375$ is less than the right side $0.5$ , which means that the given statement is always true.

True

Test a value on the interval $0 < p < \frac{27}{2}$ to see if it makes the inequality true.

Choose a value on the interval $0 < p < \frac{27}{2}$ and see if this value makes the original inequality true.

$p = 2$

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

$\frac{8}{2 (2)} + \frac{11}{4 (2)} < \frac{1}{2}$

The left side $3.375$ is not less than the right side $0.5$ , which means that the given statement is false.

False

Test a value on the interval $p > \frac{27}{2}$ to see if it makes the inequality true.

Choose a value on the interval $p > \frac{27}{2}$ and see if this value makes the original inequality true.

$p = 16$

Replace $p$ with $16$ in the original inequality.

$\frac{8}{2 (16)} + \frac{11}{4 (16)} < \frac{1}{2}$

The left side $0.421875$ 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.

$p < 0$ True

$0 < p < \frac{27}{2}$ False

$p > \frac{27}{2}$ True

$p < 0$ True

$0 < p < \frac{27}{2}$ False

$p > \frac{27}{2}$ True

Step 8

The solution consists of all of the true intervals.

$p < 0$ or $p > \frac{27}{2}$

Step 9

The result can be shown in multiple forms.

Inequality Form:

$p < 0 or p > \frac{27}{2}$

Interval Notation:

$(- \infty, 0) \cup (\frac{27}{2}, \infty)$

Step 10

Solve for p 8/(2p)+11/(4p)<1/2

Related Questions

Support LCMGCF.com by disabling your adblocker.