Precalculus Examples

$\frac{2}{w} + 3 > \frac{29}{w}$

Step 1

Multiply both sides by $w$ .

$(\frac{2}{w} + 3) w = \frac{29}{w} w$

Step 2

Simplify.

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Step 2.1

Simplify the left side.

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Step 2.1.1

Simplify $(\frac{2}{w} + 3) w$ .

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Step 2.1.1.1

Apply the distributive property.

$\frac{2}{w} w + 3 w = \frac{29}{w} w$

Step 2.1.1.2

Cancel the common factor of $w$ .

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Step 2.1.1.2.1

Cancel the common factor.

$\frac{2}{w} w + 3 w = \frac{29}{w} w$

Step 2.1.1.2.2

Rewrite the expression.

$2 + 3 w = \frac{29}{w} w$

Step 2.1.1.3

Reorder $2$ and $3 w$ .

$3 w + 2 = \frac{29}{w} w$

Step 2.2

Simplify the right side.

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Step 2.2.1

Cancel the common factor of $w$ .

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Step 2.2.1.1

Cancel the common factor.

$3 w + 2 = \frac{29}{w} w$

Step 2.2.1.2

Rewrite the expression.

$3 w + 2 = 29$

Step 3

Solve for $w$ .

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Step 3.1

Move all terms not containing $w$ to the right side of the equation.

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Step 3.1.1

Subtract $2$ from both sides of the equation.

$3 w = 29 - 2$

Step 3.1.2

Subtract $2$ from $29$ .

$3 w = 27$

Step 3.2

Divide each term in $3 w = 27$ by $3$ and simplify.

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Step 3.2.1

Divide each term in $3 w = 27$ by $3$ .

$\frac{3 w}{3} = \frac{27}{3}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $3$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{3 w}{3} = \frac{27}{3}$

Step 3.2.2.1.2

Divide $w$ by $1$ .

$w = \frac{27}{3}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Divide $27$ by $3$ .

$w = 9$

Step 4

Find the domain of $3 - \frac{27}{w}$ .

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Step 4.1

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

$w = 0$

Step 4.2

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

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

Step 5

Use each root to create test intervals.

$w < 0$

$0 < w < 9$

$w > 9$

Step 6

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 6.1

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

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Step 6.1.1

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

$w = - 2$

Step 6.1.2

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

$\frac{2}{- 2} + 3 > \frac{29}{- 2}$

Step 6.1.3

The left side $2$ is greater than the right side $- 14.5$ , which means that the given statement is always true.

True

Step 6.2

Test a value on the interval $0 < w < 9$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $0 < w < 9$ and see if this value makes the original inequality true.

$w = 4$

Step 6.2.2

Replace $w$ with $4$ in the original inequality.

$\frac{2}{4} + 3 > \frac{29}{4}$

Step 6.2.3

The left side $3.5$ is not greater than the right side $7.25$ , which means that the given statement is false.

False

Step 6.3

Test a value on the interval $w > 9$ to see if it makes the inequality true.

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Step 6.3.1

Choose a value on the interval $w > 9$ and see if this value makes the original inequality true.

$w = 12$

Step 6.3.2

Replace $w$ with $12$ in the original inequality.

$\frac{2}{12} + 3 > \frac{29}{12}$

Step 6.3.3

The left side $3.1\overline{6}$ is greater than the right side $2.41\overline{6}$ , which means that the given statement is always true.

True

Step 6.4

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

$w < 0$ True

$0 < w < 9$ False

$w > 9$ True

$w < 0$ True

$0 < w < 9$ False

$w > 9$ True

Step 7

The solution consists of all of the true intervals.

$w < 0$ or $w > 9$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$w < 0 or w > 9$

Interval Notation:

$(- \infty, 0) \cup (9, \infty)$

Step 9