Algebra Examples

$1 + \frac{5}{a - 1} \leq \frac{7}{6}$

Step 1

Subtract $\frac{7}{6}$ from both sides of the inequality.

$1 + \frac{5}{a - 1} - \frac{7}{6} \leq 0$

Step 2

Simplify $1 + \frac{5}{a - 1} - \frac{7}{6}$ .

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Step 2.1

Write $1$ as a fraction with a common denominator.

$\frac{a - 1}{a - 1} + \frac{5}{a - 1} - \frac{7}{6} \leq 0$

Step 2.2

Combine the numerators over the common denominator.

$\frac{a - 1 + 5}{a - 1} - \frac{7}{6} \leq 0$

Step 2.3

Add $- 1$ and $5$ .

$\frac{a + 4}{a - 1} - \frac{7}{6} \leq 0$

Step 2.4

To write $\frac{a + 4}{a - 1}$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$\frac{a + 4}{a - 1} \cdot \frac{6}{6} - \frac{7}{6} \leq 0$

Step 2.5

To write $- \frac{7}{6}$ as a fraction with a common denominator, multiply by $\frac{a - 1}{a - 1}$ .

$\frac{a + 4}{a - 1} \cdot \frac{6}{6} - \frac{7}{6} \cdot \frac{a - 1}{a - 1} \leq 0$

Step 2.6

Write each expression with a common denominator of $(a - 1) \cdot 6$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.6.1

Multiply $\frac{a + 4}{a - 1}$ by $\frac{6}{6}$ .

$\frac{(a + 4) \cdot 6}{(a - 1) \cdot 6} - \frac{7}{6} \cdot \frac{a - 1}{a - 1} \leq 0$

Step 2.6.2

Multiply $\frac{7}{6}$ by $\frac{a - 1}{a - 1}$ .

$\frac{(a + 4) \cdot 6}{(a - 1) \cdot 6} - \frac{7 (a - 1)}{6 (a - 1)} \leq 0$

Step 2.6.3

Reorder the factors of $(a - 1) \cdot 6$ .

$\frac{(a + 4) \cdot 6}{6 (a - 1)} - \frac{7 (a - 1)}{6 (a - 1)} \leq 0$

Step 2.7

Combine the numerators over the common denominator.

$\frac{(a + 4) \cdot 6 - 7 (a - 1)}{6 (a - 1)} \leq 0$

Step 2.8

Simplify the numerator.

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Step 2.8.1

Apply the distributive property.

$\frac{a \cdot 6 + 4 \cdot 6 - 7 (a - 1)}{6 (a - 1)} \leq 0$

Step 2.8.2

Move $6$ to the left of $a$ .

$\frac{6 \cdot a + 4 \cdot 6 - 7 (a - 1)}{6 (a - 1)} \leq 0$

Step 2.8.3

Multiply $4$ by $6$ .

$\frac{6 \cdot a + 24 - 7 (a - 1)}{6 (a - 1)} \leq 0$

Step 2.8.4

Apply the distributive property.

$\frac{6 a + 24 - 7 a - 7 \cdot - 1}{6 (a - 1)} \leq 0$

Step 2.8.5

Multiply $- 7$ by $- 1$ .

$\frac{6 a + 24 - 7 a + 7}{6 (a - 1)} \leq 0$

Step 2.8.6

Subtract $7 a$ from $6 a$ .

$\frac{- a + 24 + 7}{6 (a - 1)} \leq 0$

Step 2.8.7

Add $24$ and $7$ .

$\frac{- a + 31}{6 (a - 1)} \leq 0$

Step 2.9

Factor $- 1$ out of $- a$ .

$\frac{- (a) + 31}{6 (a - 1)} \leq 0$

Step 2.10

Rewrite $31$ as $- 1 (- 31)$ .

$\frac{- (a) - 1 (- 31)}{6 (a - 1)} \leq 0$

Step 2.11

Factor $- 1$ out of $- (a) - 1 (- 31)$ .

$\frac{- (a - 31)}{6 (a - 1)} \leq 0$

Step 2.12

Rewrite $- (a - 31)$ as $- 1 (a - 31)$ .

$\frac{- 1 (a - 31)}{6 (a - 1)} \leq 0$

Step 2.13

Move the negative in front of the fraction.

$- \frac{a - 31}{6 (a - 1)} \leq 0$

Step 3

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

$a - 31 = 0$

$a - 1 = 0$

Step 4

Add $31$ to both sides of the equation.

$a = 31$

Step 5

Add $1$ to both sides of the equation.

$a = 1$

Step 6

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

$a = 31$

$a = 1$

Step 7

Consolidate the solutions.

$a = 31, 1$

Step 8

Find the domain of $- \frac{a - 31}{6 (a - 1)}$ .

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Step 8.1

Set the denominator in $\frac{a - 31}{6 (a - 1)}$ equal to $0$ to find where the expression is undefined.

$6 (a - 1) = 0$

Step 8.2

Solve for $a$ .

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Step 8.2.1

Divide each term in $6 (a - 1) = 0$ by $6$ and simplify.

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Step 8.2.1.1

Divide each term in $6 (a - 1) = 0$ by $6$ .

$\frac{6 (a - 1)}{6} = \frac{0}{6}$

Step 8.2.1.2

Simplify the left side.

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Step 8.2.1.2.1

Cancel the common factor of $6$ .

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Step 8.2.1.2.1.1

Cancel the common factor.

$\frac{6 (a - 1)}{6} = \frac{0}{6}$

Step 8.2.1.2.1.2

Divide $a - 1$ by $1$ .

$a - 1 = \frac{0}{6}$

Step 8.2.1.3

Simplify the right side.

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Step 8.2.1.3.1

Divide $0$ by $6$ .

$a - 1 = 0$

Step 8.2.2

Add $1$ to both sides of the equation.

$a = 1$

Step 8.3

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

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

Step 9

Use each root to create test intervals.

$a < 1$

$1 < a < 31$

$a > 31$

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 $a < 1$ to see if it makes the inequality true.

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Step 10.1.1

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

$a = 0$

Step 10.1.2

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

$1 + \frac{5}{(0) - 1} \leq \frac{7}{6}$

Step 10.1.3

The left side $- 4$ is less than the right side $1.1\overline{6}$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $1 < a < 31$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $1 < a < 31$ and see if this value makes the original inequality true.

$a = 4$

Step 10.2.2

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

$1 + \frac{5}{(4) - 1} \leq \frac{7}{6}$

Step 10.2.3

The left side $2.\overline{6}$ is greater than the right side $1.1\overline{6}$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $a > 31$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $a > 31$ and see if this value makes the original inequality true.

$a = 34$

Step 10.3.2

Replace $a$ with $34$ in the original inequality.

$1 + \frac{5}{(34) - 1} \leq \frac{7}{6}$

Step 10.3.3

The left side $1.\overline{15}$ is less than the right side $1.1\overline{6}$ , which means that the given statement is always true.

True

Step 10.4

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

$a < 1$ True

$1 < a < 31$ False

$a > 31$ True

$a < 1$ True

$1 < a < 31$ False

$a > 31$ True

Step 11

The solution consists of all of the true intervals.

$a < 1$ or $a \geq 31$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$a < 1 or a \geq 31$

Interval Notation:

$(- \infty, 1) \cup [31, \infty)$

Step 13