Precalculus Examples

$\frac{7}{y + 1} > 7$

Step 1

Subtract $7$ from both sides of the inequality.

$\frac{7}{y + 1} - 7 > 0$

Step 2

Simplify $\frac{7}{y + 1} - 7$ .

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Step 2.1

To write $- 7$ as a fraction with a common denominator, multiply by $\frac{y + 1}{y + 1}$ .

$\frac{7}{y + 1} - 7 \cdot \frac{y + 1}{y + 1} > 0$

Step 2.2

Combine $- 7$ and $\frac{y + 1}{y + 1}$ .

$\frac{7}{y + 1} + \frac{- 7 (y + 1)}{y + 1} > 0$

Step 2.3

Combine the numerators over the common denominator.

$\frac{7 - 7 (y + 1)}{y + 1} > 0$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Factor $7$ out of $7 - 7 (y + 1)$ .

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Step 2.4.1.1

Factor $7$ out of $7$ .

$\frac{7 (1) - 7 (y + 1)}{y + 1} > 0$

Step 2.4.1.2

Factor $7$ out of $- 7 (y + 1)$ .

$\frac{7 (1) + 7 (- (y + 1))}{y + 1} > 0$

Step 2.4.1.3

Factor $7$ out of $7 (1) + 7 (- (y + 1))$ .

$\frac{7 (1 - (y + 1))}{y + 1} > 0$

Step 2.4.2

Apply the distributive property.

$\frac{7 (1 - y - 1 \cdot 1)}{y + 1} > 0$

Step 2.4.3

Multiply $- 1$ by $1$ .

$\frac{7 (1 - y - 1)}{y + 1} > 0$

Step 2.4.4

Subtract $1$ from $1$ .

$\frac{7 (- y + 0)}{y + 1} > 0$

Step 2.4.5

Add $- y$ and $0$ .

$\frac{7 \cdot - 1 y}{y + 1} > 0$

Step 2.4.6

Combine exponents.

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Step 2.4.6.1

Factor out negative.

$\frac{- (7 y)}{y + 1} > 0$

Step 2.4.6.2

Multiply $7$ by $- 1$ .

$\frac{- 7 y}{y + 1} > 0$

Step 2.5

Move the negative in front of the fraction.

$- \frac{7 y}{y + 1} > 0$

Step 3

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

$- 7 y = 0$

$y + 1 = 0$

Step 4

Divide each term in $- 7 y = 0$ by $- 7$ and simplify.

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Step 4.1

Divide each term in $- 7 y = 0$ by $- 7$ .

$\frac{- 7 y}{- 7} = \frac{0}{- 7}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 7$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 7 y}{- 7} = \frac{0}{- 7}$

Step 4.2.1.2

Divide $y$ by $1$ .

$y = \frac{0}{- 7}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $- 7$ .

$y = 0$

Step 5

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 6

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

$y = 0$

$y = - 1$

Step 7

Consolidate the solutions.

$y = 0, - 1$

Step 8

Find the domain of $- \frac{7 y}{y + 1}$ .

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Step 8.1

Set the denominator in $\frac{7 y}{y + 1}$ equal to $0$ to find where the expression is undefined.

$y + 1 = 0$

Step 8.2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 8.3

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

$(- \infty, - 1) \cup (- 1, \infty)$

Step 9

Use each root to create test intervals.

$y < - 1$

$- 1 < y < 0$

$y > 0$

Step 10

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 10.1

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

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Step 10.1.1

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

$y = - 4$

Step 10.1.2

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

$\frac{7}{(- 4) + 1} > 7$

Step 10.1.3

The left side $- 2.\overline{3}$ is not greater than the right side $7$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $- 1 < y < 0$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 1 < y < 0$ and see if this value makes the original inequality true.

$y = - 0.5$

Step 10.2.2

Replace $y$ with $- 0.5$ in the original inequality.

$\frac{7}{(- 0.5) + 1} > 7$

Step 10.2.3

The left side $14$ is greater than the right side $7$ , which means that the given statement is always true.

True

Step 10.3

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

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Step 10.3.1

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

$y = 2$

Step 10.3.2

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

$\frac{7}{(2) + 1} > 7$

Step 10.3.3

The left side $2.\overline{3}$ is not greater than the right side $7$ , which means that the given statement is false.

False

Step 10.4

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

$y < - 1$ False

$- 1 < y < 0$ True

$y > 0$ False

$y < - 1$ False

$- 1 < y < 0$ True

$y > 0$ False

Step 11

The solution consists of all of the true intervals.

$- 1 < y < 0$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$- 1 < y < 0$

Interval Notation:

$(- 1, 0)$

Step 13