Algebra Examples

$\frac{2 x + 1}{x - 5} \leq 3$

Step 1

Subtract $3$ from both sides of the inequality.

$\frac{2 x + 1}{x - 5} - 3 \leq 0$

Step 2

Simplify $\frac{2 x + 1}{x - 5} - 3$ .

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Step 2.1

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{x - 5}{x - 5}$ .

$\frac{2 x + 1}{x - 5} - 3 \cdot \frac{x - 5}{x - 5} \leq 0$

Step 2.2

Combine $- 3$ and $\frac{x - 5}{x - 5}$ .

$\frac{2 x + 1}{x - 5} + \frac{- 3 (x - 5)}{x - 5} \leq 0$

Step 2.3

Combine the numerators over the common denominator.

$\frac{2 x + 1 - 3 (x - 5)}{x - 5} \leq 0$

Step 2.4

Simplify the numerator.

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Step 2.4.1

Apply the distributive property.

$\frac{2 x + 1 - 3 x - 3 \cdot - 5}{x - 5} \leq 0$

Step 2.4.2

Multiply $- 3$ by $- 5$ .

$\frac{2 x + 1 - 3 x + 15}{x - 5} \leq 0$

Step 2.4.3

Subtract $3 x$ from $2 x$ .

$\frac{- x + 1 + 15}{x - 5} \leq 0$

Step 2.4.4

Add $1$ and $15$ .

$\frac{- x + 16}{x - 5} \leq 0$

Step 2.5

Factor $- 1$ out of $- x$ .

$\frac{- (x) + 16}{x - 5} \leq 0$

Step 2.6

Rewrite $16$ as $- 1 (- 16)$ .

$\frac{- (x) - 1 (- 16)}{x - 5} \leq 0$

Step 2.7

Factor $- 1$ out of $- (x) - 1 (- 16)$ .

$\frac{- (x - 16)}{x - 5} \leq 0$

Step 2.8

Rewrite $- (x - 16)$ as $- 1 (x - 16)$ .

$\frac{- 1 (x - 16)}{x - 5} \leq 0$

Step 2.9

Move the negative in front of the fraction.

$- \frac{x - 16}{x - 5} \leq 0$

Step 3

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

$x - 16 = 0$

$x - 5 = 0$

Step 4

Add $16$ to both sides of the equation.

$x = 16$

Step 5

Add $5$ to both sides of the equation.

$x = 5$

Step 6

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

$x = 16$

$x = 5$

Step 7

Consolidate the solutions.

$x = 16, 5$

Step 8

Find the domain of $- \frac{x - 16}{x - 5}$ .

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Step 8.1

Set the denominator in $\frac{x - 16}{x - 5}$ equal to $0$ to find where the expression is undefined.

$x - 5 = 0$

Step 8.2

Add $5$ to both sides of the equation.

$x = 5$

Step 8.3

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

$(- \infty, 5) \cup (5, \infty)$

Step 9

Use each root to create test intervals.

$x < 5$

$5 < x < 16$

$x > 16$

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 $x < 5$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < 5$ and see if this value makes the original inequality true.

$x = 0$

Step 10.1.2

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

$\frac{2 (0) + 1}{(0) - 5} \leq 3$

Step 10.1.3

The left side $- 0.2$ is less than the right side $3$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $5 < x < 16$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $5 < x < 16$ and see if this value makes the original inequality true.

$x = 8$

Step 10.2.2

Replace $x$ with $8$ in the original inequality.

$\frac{2 (8) + 1}{(8) - 5} \leq 3$

Step 10.2.3

The left side $5.\overline{6}$ is greater than the right side $3$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $x > 16$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $x > 16$ and see if this value makes the original inequality true.

$x = 18$

Step 10.3.2

Replace $x$ with $18$ in the original inequality.

$\frac{2 (18) + 1}{(18) - 5} \leq 3$

Step 10.3.3

The left side $2.\overline{846153}$ is less than the right side $3$ , which means that the given statement is always true.

True

Step 10.4

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

$x < 5$ True

$5 < x < 16$ False

$x > 16$ True

$x < 5$ True

$5 < x < 16$ False

$x > 16$ True

Step 11

The solution consists of all of the true intervals.

$x < 5$ or $x \geq 16$

Step 12

Convert the inequality to interval notation.

$(- \infty, 5) \cup [16, \infty)$

Step 13