Precalculus Examples

$\frac{4 x - 2}{x + 3} \leq 3$

Step 1

Subtract $3$ from both sides of the inequality.

$\frac{4 x - 2}{x + 3} - 3 \leq 0$

Step 2

Simplify $\frac{4 x - 2}{x + 3} - 3$ .

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Step 2.1

Factor $2$ out of $4 x - 2$ .

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Step 2.1.1

Factor $2$ out of $4 x$ .

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

Step 2.1.2

Factor $2$ out of $- 2$ .

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

Step 2.1.3

Factor $2$ out of $2 (2 x) + 2 (- 1)$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

Combine the numerators over the common denominator.

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

Step 2.5

Simplify the numerator.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Multiply $2$ by $2$ .

$\frac{4 x + 2 \cdot - 1 - 3 (x + 3)}{x + 3} \leq 0$

Step 2.5.3

Multiply $2$ by $- 1$ .

$\frac{4 x - 2 - 3 (x + 3)}{x + 3} \leq 0$

Step 2.5.4

Apply the distributive property.

$\frac{4 x - 2 - 3 x - 3 \cdot 3}{x + 3} \leq 0$

Step 2.5.5

Multiply $- 3$ by $3$ .

$\frac{4 x - 2 - 3 x - 9}{x + 3} \leq 0$

Step 2.5.6

Subtract $3 x$ from $4 x$ .

$\frac{x - 2 - 9}{x + 3} \leq 0$

Step 2.5.7

Subtract $9$ from $- 2$ .

$\frac{x - 11}{x + 3} \leq 0$

Step 3

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

$x - 11 = 0$

$x + 3 = 0$

Step 4

Add $11$ to both sides of the equation.

$x = 11$

Step 5

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

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

$x = 11$

$x = - 3$

Step 7

Consolidate the solutions.

$x = 11, - 3$

Step 8

Find the domain of $\frac{x - 11}{x + 3}$ .

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Step 8.1

Set the denominator in $\frac{x - 11}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Step 8.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 8.3

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

$(- \infty, - 3) \cup (- 3, \infty)$

Step 9

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 11$

$x > 11$

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 $x < - 3$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Step 10.1.2

Replace $x$ with $- 6$ in the original inequality.

$\frac{4 (- 6) - 2}{(- 6) + 3} \leq 3$

Step 10.1.3

The left side $8.\overline{6}$ is greater than the right side $3$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $- 3 < x < 11$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 3 < x < 11$ and see if this value makes the original inequality true.

$x = 0$

Step 10.2.2

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

$\frac{4 (0) - 2}{(0) + 3} \leq 3$

Step 10.2.3

The left side $- 0.\overline{6}$ is less than the right side $3$ , which means that the given statement is always true.

True

Step 10.3

Test a value on the interval $x > 11$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $x > 11$ and see if this value makes the original inequality true.

$x = 14$

Step 10.3.2

Replace $x$ with $14$ in the original inequality.

$\frac{4 (14) - 2}{(14) + 3} \leq 3$

Step 10.3.3

The left side $3.17647058$ is greater than the right side $3$ , which means that the given statement is false.

False

Step 10.4

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

$x < - 3$ False

$- 3 < x < 11$ True

$x > 11$ False

$x < - 3$ False

$- 3 < x < 11$ True

$x > 11$ False

Step 11

The solution consists of all of the true intervals.

$- 3 < x \leq 11$

Step 12

Convert the inequality to interval notation.

$(- 3, 11]$

Step 13