Precalculus Examples

$\frac{1}{2 b + 1} + \frac{1}{b + 1} > \frac{8}{15}$

Step 1

Subtract $\frac{8}{15}$ from both sides of the inequality.

$\frac{1}{2 b + 1} + \frac{1}{b + 1} - \frac{8}{15} > 0$

Step 2

Simplify $\frac{1}{2 b + 1} + \frac{1}{b + 1} - \frac{8}{15}$ .

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Step 2.1

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

$\frac{1}{2 b + 1} \cdot \frac{b + 1}{b + 1} + \frac{1}{b + 1} - \frac{8}{15} > 0$

Step 2.2

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

$\frac{1}{2 b + 1} \cdot \frac{b + 1}{b + 1} + \frac{1}{b + 1} \cdot \frac{2 b + 1}{2 b + 1} - \frac{8}{15} > 0$

Step 2.3

Write each expression with a common denominator of $(2 b + 1) (b + 1)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.1

Multiply $\frac{1}{2 b + 1}$ by $\frac{b + 1}{b + 1}$ .

$\frac{b + 1}{(2 b + 1) (b + 1)} + \frac{1}{b + 1} \cdot \frac{2 b + 1}{2 b + 1} - \frac{8}{15} > 0$

Step 2.3.2

Multiply $\frac{1}{b + 1}$ by $\frac{2 b + 1}{2 b + 1}$ .

$\frac{b + 1}{(2 b + 1) (b + 1)} + \frac{2 b + 1}{(b + 1) (2 b + 1)} - \frac{8}{15} > 0$

Step 2.3.3

Reorder the factors of $(b + 1) (2 b + 1)$ .

$\frac{b + 1}{(2 b + 1) (b + 1)} + \frac{2 b + 1}{(2 b + 1) (b + 1)} - \frac{8}{15} > 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{b + 1 + 2 b + 1}{(2 b + 1) (b + 1)} - \frac{8}{15} > 0$

Step 2.5

Add $b$ and $2 b$ .

$\frac{3 b + 1 + 1}{(2 b + 1) (b + 1)} - \frac{8}{15} > 0$

Step 2.6

Add $1$ and $1$ .

$\frac{3 b + 2}{(2 b + 1) (b + 1)} - \frac{8}{15} > 0$

Step 2.7

To write $\frac{3 b + 2}{(2 b + 1) (b + 1)}$ as a fraction with a common denominator, multiply by $\frac{15}{15}$ .

$\frac{3 b + 2}{(2 b + 1) (b + 1)} \cdot \frac{15}{15} - \frac{8}{15} > 0$

Step 2.8

To write $- \frac{8}{15}$ as a fraction with a common denominator, multiply by $\frac{(2 b + 1) (b + 1)}{(2 b + 1) (b + 1)}$ .

$\frac{3 b + 2}{(2 b + 1) (b + 1)} \cdot \frac{15}{15} - \frac{8}{15} \cdot \frac{(2 b + 1) (b + 1)}{(2 b + 1) (b + 1)} > 0$

Step 2.9

Write each expression with a common denominator of $(2 b + 1) (b + 1) \cdot 15$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.9.1

Multiply $\frac{3 b + 2}{(2 b + 1) (b + 1)}$ by $\frac{15}{15}$ .

$\frac{(3 b + 2) \cdot 15}{(2 b + 1) (b + 1) \cdot 15} - \frac{8}{15} \cdot \frac{(2 b + 1) (b + 1)}{(2 b + 1) (b + 1)} > 0$

Step 2.9.2

Multiply $\frac{8}{15}$ by $\frac{(2 b + 1) (b + 1)}{(2 b + 1) (b + 1)}$ .

$\frac{(3 b + 2) \cdot 15}{(2 b + 1) (b + 1) \cdot 15} - \frac{8 ((2 b + 1) (b + 1))}{15 ((2 b + 1) (b + 1))} > 0$

Step 2.9.3

Reorder the factors of $(2 b + 1) (b + 1) \cdot 15$ .

$\frac{(3 b + 2) \cdot 15}{15 (2 b + 1) (b + 1)} - \frac{8 ((2 b + 1) (b + 1))}{15 ((2 b + 1) (b + 1))} > 0$

Step 2.9.4

Reorder the factors of $15 ((2 b + 1) (b + 1))$ .

$\frac{(3 b + 2) \cdot 15}{15 (2 b + 1) (b + 1)} - \frac{8 ((2 b + 1) (b + 1))}{15 (2 b + 1) (b + 1)} > 0$

Step 2.10

Combine the numerators over the common denominator.

$\frac{(3 b + 2) \cdot 15 - 8 ((2 b + 1) (b + 1))}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11

Simplify the numerator.

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Step 2.11.1

Apply the distributive property.

$\frac{3 b \cdot 15 + 2 \cdot 15 - 8 (2 b + 1) (b + 1)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.2

Multiply $15$ by $3$ .

$\frac{45 b + 2 \cdot 15 - 8 (2 b + 1) (b + 1)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.3

Multiply $2$ by $15$ .

$\frac{45 b + 30 - 8 (2 b + 1) (b + 1)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.4

Apply the distributive property.

$\frac{45 b + 30 + (- 8 (2 b) - 8 \cdot 1) (b + 1)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.5

Multiply $2$ by $- 8$ .

$\frac{45 b + 30 + (- 16 b - 8 \cdot 1) (b + 1)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.6

Multiply $- 8$ by $1$ .

$\frac{45 b + 30 + (- 16 b - 8) (b + 1)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.7

Expand $(- 16 b - 8) (b + 1)$ using the FOIL Method.

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Step 2.11.7.1

Apply the distributive property.

$\frac{45 b + 30 - 16 b (b + 1) - 8 (b + 1)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.7.2

Apply the distributive property.

$\frac{45 b + 30 - 16 b \cdot b - 16 b \cdot 1 - 8 (b + 1)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.7.3

Apply the distributive property.

$\frac{45 b + 30 - 16 b \cdot b - 16 b \cdot 1 - 8 b - 8 \cdot 1}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.8

Simplify and combine like terms.

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Step 2.11.8.1

Simplify each term.

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Step 2.11.8.1.1

Multiply $b$ by $b$ by adding the exponents.

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Step 2.11.8.1.1.1

Move $b$ .

$\frac{45 b + 30 - 16 (b \cdot b) - 16 b \cdot 1 - 8 b - 8 \cdot 1}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.8.1.1.2

Multiply $b$ by $b$ .

$\frac{45 b + 30 - 16 b^{2} - 16 b \cdot 1 - 8 b - 8 \cdot 1}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.8.1.2

Multiply $- 16$ by $1$ .

$\frac{45 b + 30 - 16 b^{2} - 16 b - 8 b - 8 \cdot 1}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.8.1.3

Multiply $- 8$ by $1$ .

$\frac{45 b + 30 - 16 b^{2} - 16 b - 8 b - 8}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.8.2

Subtract $8 b$ from $- 16 b$ .

$\frac{45 b + 30 - 16 b^{2} - 24 b - 8}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.9

Subtract $24 b$ from $45 b$ .

$\frac{21 b + 30 - 16 b^{2} - 8}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.10

Subtract $8$ from $30$ .

$\frac{21 b - 16 b^{2} + 22}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.11

Reorder terms.

$\frac{- 16 b^{2} + 21 b + 22}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.12

Factor by grouping.

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Step 2.11.12.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 16 \cdot 22 = - 352$ and whose sum is $b = 21$ .

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Step 2.11.12.1.1

Factor $21$ out of $21 b$ .

$\frac{- 16 b^{2} + 21 (b) + 22}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.12.1.2

Rewrite $21$ as $- 11$ plus $32$

$\frac{- 16 b^{2} + (- 11 + 32) b + 22}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.12.1.3

Apply the distributive property.

$\frac{- 16 b^{2} - 11 b + 32 b + 22}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.12.2

Factor out the greatest common factor from each group.

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Step 2.11.12.2.1

Group the first two terms and the last two terms.

$\frac{(- 16 b^{2} - 11 b) + 32 b + 22}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.12.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{b (- 16 b - 11) - 2 (- 16 b - 11)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.11.12.3

Factor the polynomial by factoring out the greatest common factor, $- 16 b - 11$ .

$\frac{(- 16 b - 11) (b - 2)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.12

Factor $- 1$ out of $- 16 b$ .

$\frac{(- (16 b) - 11) (b - 2)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.13

Rewrite $- 11$ as $- 1 (11)$ .

$\frac{(- (16 b) - 1 (11)) (b - 2)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.14

Factor $- 1$ out of $- (16 b) - 1 (11)$ .

$\frac{- (16 b + 11) (b - 2)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.15

Rewrite $- (16 b + 11)$ as $- 1 (16 b + 11)$ .

$\frac{- 1 (16 b + 11) (b - 2)}{15 (2 b + 1) (b + 1)} > 0$

Step 2.16

Move the negative in front of the fraction.

$- \frac{(16 b + 11) (b - 2)}{15 (2 b + 1) (b + 1)} > 0$

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$16 b + 11 = 0$

$b - 2 = 0$

$2 b + 1 = 0$

$b + 1 = 0$

Step 4

Subtract $11$ from both sides of the equation.

$16 b = - 11$

Step 5

Divide each term in $16 b = - 11$ by $16$ and simplify.

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Step 5.1

Divide each term in $16 b = - 11$ by $16$ .

$\frac{16 b}{16} = \frac{- 11}{16}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $16$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{16 b}{16} = \frac{- 11}{16}$

Step 5.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 11}{16}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Move the negative in front of the fraction.

$b = - \frac{11}{16}$

Step 6

Add $2$ to both sides of the equation.

$b = 2$

Step 7

Subtract $1$ from both sides of the equation.

$2 b = - 1$

Step 8

Divide each term in $2 b = - 1$ by $2$ and simplify.

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Step 8.1

Divide each term in $2 b = - 1$ by $2$ .

$\frac{2 b}{2} = \frac{- 1}{2}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $2$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{2 b}{2} = \frac{- 1}{2}$

Step 8.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 1}{2}$

Step 8.3

Simplify the right side.

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Step 8.3.1

Move the negative in front of the fraction.

$b = - \frac{1}{2}$

Step 9

Subtract $1$ from both sides of the equation.

$b = - 1$

Step 10

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$b = - \frac{11}{16}$

$b = 2$

$b = - \frac{1}{2}$

$b = - 1$

Step 11

Consolidate the solutions.

$b = - \frac{11}{16}, 2, - \frac{1}{2}, - 1$

Step 12

Find the domain of $- \frac{(16 b + 11) (b - 2)}{15 (2 b + 1) (b + 1)}$ .

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Step 12.1

Set the denominator in $\frac{(16 b + 11) (b - 2)}{15 (2 b + 1) (b + 1)}$ equal to $0$ to find where the expression is undefined.

$15 (2 b + 1) (b + 1) = 0$

Step 12.2

Solve for $b$ .

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Step 12.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 b + 1 = 0$

$b + 1 = 0$

Step 12.2.2

Set $2 b + 1$ equal to $0$ and solve for $b$ .

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Step 12.2.2.1

Set $2 b + 1$ equal to $0$ .

$2 b + 1 = 0$

Step 12.2.2.2

Solve $2 b + 1 = 0$ for $b$ .

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Step 12.2.2.2.1

Subtract $1$ from both sides of the equation.

$2 b = - 1$

Step 12.2.2.2.2

Divide each term in $2 b = - 1$ by $2$ and simplify.

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Step 12.2.2.2.2.1

Divide each term in $2 b = - 1$ by $2$ .

$\frac{2 b}{2} = \frac{- 1}{2}$

Step 12.2.2.2.2.2

Simplify the left side.

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Step 12.2.2.2.2.2.1

Cancel the common factor of $2$ .

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Step 12.2.2.2.2.2.1.1

Cancel the common factor.

$\frac{2 b}{2} = \frac{- 1}{2}$

Step 12.2.2.2.2.2.1.2

Divide $b$ by $1$ .

$b = \frac{- 1}{2}$

Step 12.2.2.2.2.3

Simplify the right side.

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Step 12.2.2.2.2.3.1

Move the negative in front of the fraction.

$b = - \frac{1}{2}$

Step 12.2.3

Set $b + 1$ equal to $0$ and solve for $b$ .

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Step 12.2.3.1

Set $b + 1$ equal to $0$ .

$b + 1 = 0$

Step 12.2.3.2

Subtract $1$ from both sides of the equation.

$b = - 1$

Step 12.2.4

The final solution is all the values that make $15 (2 b + 1) (b + 1) = 0$ true.

$b = - \frac{1}{2}, - 1$

Step 12.3

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

$(- \infty, - 1) \cup (- 1, - \frac{1}{2}) \cup (- \frac{1}{2}, \infty)$

Step 13

Use each root to create test intervals.

$b < - 1$

$- 1 < b < - \frac{11}{16}$

$- \frac{11}{16} < b < - \frac{1}{2}$

$- \frac{1}{2} < b < 2$

$b > 2$

Step 14

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 14.1

Test a value on the interval $b < - 1$ to see if it makes the inequality true.

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Step 14.1.1

Choose a value on the interval $b < - 1$ and see if this value makes the original inequality true.

$b = - 4$

Step 14.1.2

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

$\frac{1}{2 (- 4) + 1} + \frac{1}{(- 4) + 1} > \frac{8}{15}$

Step 14.1.3

The left side $- 0.\overline{476190}$ is not greater than the right side $0.5\overline{3}$ , which means that the given statement is false.

False

Step 14.2

Test a value on the interval $- 1 < b < - \frac{11}{16}$ to see if it makes the inequality true.

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Step 14.2.1

Choose a value on the interval $- 1 < b < - \frac{11}{16}$ and see if this value makes the original inequality true.

$b = - 0.84$

Step 14.2.2

Replace $b$ with $- 0.84$ in the original inequality.

$\frac{1}{2 (- 0.84) + 1} + \frac{1}{(- 0.84) + 1} > \frac{8}{15}$

Step 14.2.3

The left side $4.77941176$ is greater than the right side $0.5\overline{3}$ , which means that the given statement is always true.

True

Step 14.3

Test a value on the interval $- \frac{11}{16} < b < - \frac{1}{2}$ to see if it makes the inequality true.

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Step 14.3.1

Choose a value on the interval $- \frac{11}{16} < b < - \frac{1}{2}$ and see if this value makes the original inequality true.

$b = - 0.59$

Step 14.3.2

Replace $b$ with $- 0.59$ in the original inequality.

$\frac{1}{2 (- 0.59) + 1} + \frac{1}{(- 0.59) + 1} > \frac{8}{15}$

Step 14.3.3

The left side $- 3.\overline{11653}$ is not greater than the right side $0.5\overline{3}$ , which means that the given statement is false.

False

Step 14.4

Test a value on the interval $- \frac{1}{2} < b < 2$ to see if it makes the inequality true.

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Step 14.4.1

Choose a value on the interval $- \frac{1}{2} < b < 2$ and see if this value makes the original inequality true.

$b = 0$

Step 14.4.2

Replace $b$ with $0$ in the original inequality.

$\frac{1}{2 (0) + 1} + \frac{1}{(0) + 1} > \frac{8}{15}$

Step 14.4.3

The left side $2$ is greater than the right side $0.5\overline{3}$ , which means that the given statement is always true.

True

Step 14.5

Test a value on the interval $b > 2$ to see if it makes the inequality true.

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Step 14.5.1

Choose a value on the interval $b > 2$ and see if this value makes the original inequality true.

$b = 4$

Step 14.5.2

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

$\frac{1}{2 (4) + 1} + \frac{1}{(4) + 1} > \frac{8}{15}$

Step 14.5.3

The left side $0.3\overline{1}$ is not greater than the right side $0.5\overline{3}$ , which means that the given statement is false.

False

Step 14.6

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

$b < - 1$ False

$- 1 < b < - \frac{11}{16}$ True

$- \frac{11}{16} < b < - \frac{1}{2}$ False

$- \frac{1}{2} < b < 2$ True

$b > 2$ False

$b < - 1$ False

$- 1 < b < - \frac{11}{16}$ True

$- \frac{11}{16} < b < - \frac{1}{2}$ False

$- \frac{1}{2} < b < 2$ True

$b > 2$ False

Step 15

The solution consists of all of the true intervals.

$- 1 < b < - \frac{11}{16}$ or $- \frac{1}{2} < b < 2$

Step 16

The result can be shown in multiple forms.

Inequality Form:

$- 1 < b < - \frac{11}{16} or - \frac{1}{2} < b < 2$

Interval Notation:

$(- 1, - \frac{11}{16}) \cup (- \frac{1}{2}, 2)$

Step 17