Algebra Examples

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

Step 1

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

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Step 1.1

To write $\frac{1}{x + 1}$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

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

Step 1.2

To write $\frac{1}{x - 3}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

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

Step 1.3

Write each expression with a common denominator of $(x + 1) (x - 3)$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.3.1

Multiply $\frac{1}{x + 1}$ by $\frac{x - 3}{x - 3}$ .

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

Step 1.3.2

Multiply $\frac{1}{x - 3}$ by $\frac{x + 1}{x + 1}$ .

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

Step 1.3.3

Reorder the factors of $(x - 3) (x + 1)$ .

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

Step 1.4

Combine the numerators over the common denominator.

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

Step 1.5

Simplify the numerator.

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Step 1.5.1

Add $x$ and $x$ .

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

Step 1.5.2

Add $- 3$ and $1$ .

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

Step 1.5.3

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

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Step 1.5.3.1

Factor $2$ out of $2 x$ .

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

Step 1.5.3.2

Factor $2$ out of $- 2$ .

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

Step 1.5.3.3

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

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

Step 2

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

$x - 1 = 0$

$x + 1 = 0$

$x - 3 = 0$

Step 3

Add $1$ to both sides of the equation.

$x = 1$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

Add $3$ to 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.

$x = 1$

$x = - 1$

$x = 3$

Step 7

Consolidate the solutions.

$x = 1, - 1, 3$

Step 8

Find the domain of $\frac{1}{x + 1} + \frac{1}{x - 3}$ .

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Step 8.1

Set the denominator in $\frac{1}{x + 1}$ 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

Set the denominator in $\frac{1}{x - 3}$ equal to $0$ to find where the expression is undefined.

$x - 3 = 0$

Step 8.4

Add $3$ to both sides of the equation.

$x = 3$

Step 8.5

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

$(- \infty, - 1) \cup (- 1, 3) \cup (3, \infty)$

Step 9

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$1 < x < 3$

$x > 3$

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 $x < - 1$ 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.

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

Step 10.1.3

The left side $- 0.\overline{476190}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 10.2

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

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Step 10.2.1

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

$x = 0$

Step 10.2.2

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

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

Step 10.2.3

The left side $0.\overline{6}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $1 < x < 3$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $1 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Step 10.3.2

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

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

Step 10.3.3

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

True

Step 10.4

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

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Step 10.4.1

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

$x = 6$

Step 10.4.2

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

$\frac{1}{(6) + 1} + \frac{1}{(6) - 3} \leq 0$

Step 10.4.3

The left side $0.\overline{476190}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.5

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

$x < - 1$ True

$- 1 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

$x < - 1$ True

$- 1 < x < 1$ False

$1 < x < 3$ True

$x > 3$ False

Step 11

The solution consists of all of the true intervals.

$x < - 1$ or $1 \leq x < 3$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or 1 \leq x < 3$

Interval Notation:

$(- \infty, - 1) \cup [1, 3)$

Step 13