Algebra Examples

$\frac{2 x + 2}{x + 3} > 4$

Step 1

Subtract $4$ from both sides of the inequality.

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

Step 2

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

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Step 2.1

Factor $2$ out of $2 x + 2$ .

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Step 2.1.1

Factor $2$ out of $2 x$ .

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

Step 2.1.2

Factor $2$ out of $2$ .

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

Step 2.1.3

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

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

Step 2.2

To write $- 4$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

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

Step 2.3

Combine $- 4$ and $\frac{x + 3}{x + 3}$ .

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

Step 2.4

Combine the numerators over the common denominator.

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

Step 2.5

Simplify the numerator.

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Step 2.5.1

Factor $2$ out of $2 (x + 1) - 4 (x + 3)$ .

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Step 2.5.1.1

Factor $2$ out of $- 4 (x + 3)$ .

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

Step 2.5.1.2

Factor $2$ out of $2 (x + 1) + 2 (- 2 (x + 3))$ .

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

Step 2.5.2

Apply the distributive property.

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

Step 2.5.3

Multiply $- 2$ by $3$ .

$\frac{2 (x + 1 - 2 x - 6)}{x + 3} > 0$

Step 2.5.4

Subtract $2 x$ from $x$ .

$\frac{2 (- x + 1 - 6)}{x + 3} > 0$

Step 2.5.5

Subtract $6$ from $1$ .

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

Step 2.6

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

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

Step 2.7

Rewrite $- 5$ as $- 1 (5)$ .

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

Step 2.8

Factor $- 1$ out of $- (x) - 1 (5)$ .

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

Step 2.9

Rewrite $- (x + 5)$ as $- 1 (x + 5)$ .

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

Step 2.10

Move the negative in front of the fraction.

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

Step 3

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

$x + 5 = 0$

$x + 3 = 0$

Step 4

Subtract $5$ from both sides of the equation.

$x = - 5$

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.

$x = - 5$

$x = - 3$

Step 7

Consolidate the solutions.

$x = - 5, - 3$

Step 8

Find the domain of $- \frac{2 (x + 5)}{x + 3}$ .

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Step 8.1

Set the denominator in $\frac{2 (x + 5)}{x + 3}$ 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.

$x < - 5$

$- 5 < x < - 3$

$x > - 3$

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.

$x = - 8$

Step 10.1.2

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

$\frac{2 (- 8) + 2}{(- 8) + 3} > 4$

Step 10.1.3

The left side $2.8$ is not greater than the right side $4$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $- 5 < x < - 3$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 5 < x < - 3$ and see if this value makes the original inequality true.

$x = - 4$

Step 10.2.2

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

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

Step 10.2.3

The left side $6$ is greater than the right side $4$ , which means that the given statement is always true.

True

Step 10.3

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

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Step 10.3.1

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

$x = 0$

Step 10.3.2

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

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

Step 10.3.3

The left side $0.\overline{6}$ is not greater than the right side $4$ , which means that the given statement is false.

False

Step 10.4

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

$x < - 5$ False

$- 5 < x < - 3$ True

$x > - 3$ False

$x < - 5$ False

$- 5 < x < - 3$ True

$x > - 3$ False

Step 11

The solution consists of all of the true intervals.

$- 5 < x < - 3$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$- 5 < x < - 3$

Interval Notation:

$(- 5, - 3)$

Step 13