Algebra Examples

$\frac{4 x - 2}{x + 1} \geq - 2$

Step 1

Add $2$ to both sides of the inequality.

$\frac{4 x - 2}{x + 1} + 2 \geq 0$

Step 2

Simplify $\frac{4 x - 2}{x + 1} + 2$ .

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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 $4 x$ .

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

Step 2.1.2

Factor $2$ out of $- 2$ .

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

Step 2.1.3

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

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

Step 2.2

To write $2$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

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

Step 2.3

Combine the numerators over the common denominator.

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

Step 2.4

Simplify the numerator.

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Step 2.4.1

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

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

Step 2.4.2

Add $2 x$ and $x$ .

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

Step 2.4.3

Add $- 1$ and $1$ .

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

Step 2.4.4

Add $3 x$ and $0$ .

$\frac{2 \cdot 3 x}{x + 1} \geq 0$

Step 2.4.5

Multiply $2$ by $3$ .

$\frac{6 x}{x + 1} \geq 0$

Step 3

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

$6 x = 0$

$x + 1 = 0$

Step 4

Divide each term in $6 x = 0$ by $6$ and simplify.

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Step 4.1

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $6$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $6$ .

$x = 0$

Step 5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

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

$x = 0$

$x = - 1$

Step 7

Consolidate the solutions.

$x = 0, - 1$

Step 8

Find the domain of $\frac{6 x}{x + 1}$ .

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Step 8.1

Set the denominator in $\frac{6 x}{x + 1}$ equal to $0$ to find where the expression is undefined.

$x + 1 = 0$

Step 8.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 8.3

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

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

Step 9

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$x > 0$

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

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Step 10.1.1

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

$x = - 4$

Step 10.1.2

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

$\frac{4 (- 4) - 2}{(- 4) + 1} \geq - 2$

Step 10.1.3

The left side $6$ is greater than the right side $- 2$ , which means that the given statement is always true.

True

Step 10.2

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

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Step 10.2.1

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

$x = - 0.5$

Step 10.2.2

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

$\frac{4 (- 0.5) - 2}{(- 0.5) + 1} \geq - 2$

Step 10.2.3

The left side $- 8$ is less than the right side $- 2$ , which means that the given statement is false.

False

Step 10.3

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

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Step 10.3.1

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

$x = 2$

Step 10.3.2

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

$\frac{4 (2) - 2}{(2) + 1} \geq - 2$

Step 10.3.3

The left side $2$ is greater than the right side $- 2$ , which means that the given statement is always true.

True

Step 10.4

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

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ True

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ True

Step 11

The solution consists of all of the true intervals.

$x < - 1$ or $x \geq 0$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or x \geq 0$

Interval Notation:

$(- \infty, - 1) \cup [0, \infty)$

Step 13