Precalculus Examples

$\frac{x - 3}{x + 1} < 1$

Step 1

Subtract $1$ from both sides of the inequality.

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Combine $- 1$ and $\frac{x + 1}{x + 1}$ .

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

Step 2.3

Combine the numerators over the common denominator.

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

Step 2.4

Simplify the numerator.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Multiply $- 1$ by $1$ .

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

Step 2.4.3

Subtract $x$ from $x$ .

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

Step 2.4.4

Subtract $3$ from $0$ .

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

Step 2.4.5

Subtract $1$ from $- 3$ .

$\frac{- 4}{x + 1} < 0$

Step 2.5

Move the negative in front of the fraction.

$- \frac{4}{x + 1} < 0$

Step 3

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

$x + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

Find the domain of $- \frac{4}{x + 1}$ .

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Step 5.1

Set the denominator in $\frac{4}{x + 1}$ equal to $0$ to find where the expression is undefined.

$x + 1 = 0$

Step 5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5.3

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

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

Step 6

The solution consists of all of the true intervals.

$x > - 1$

Step 7

Convert the inequality to interval notation.

$(- 1, \infty)$

Step 8