Solve the Inequality for x (x^2+2x+3)/(x+1)<=x+2

$\frac{x^{2} + 2 x + 3}{x + 1} \leq x + 2$

Move all the expressions to the left side of the equation.

Subtract $x$ from both sides of the inequality.

$\frac{x^{2} + 2 x + 3}{x + 1} - x \leq 2$

Subtract $2$ from both sides of the inequality.

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

Simplify $\frac{x^{2} + 2 x + 3}{x + 1} - x - 2$ .

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

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

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $1$ .

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

Subtract $x^{2}$ from $x^{2}$ .

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

Add $2 x$ and $0$ .

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

Subtract $x$ from $2 x$ .

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

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

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

Combine $- 2$ and $\frac{x + 1}{x + 1}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 2$ by $1$ .

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

Subtract $2 x$ from $x$ .

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

Subtract $2$ from $3$ .

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

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

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

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

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

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

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

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

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

Move the negative in front of the fraction.

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

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

$x - 1 = 0$

$x + 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = 1$

$x = - 1$

Consolidate the solutions.

$x = 1, - 1$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$x > 1$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = - 4$

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

$\frac{{(- 4)}^{2} + 2 (- 4) + 3}{(- 4) + 1} \leq (- 4) + 2$

The left side $- 3.\overline{6}$ is less than the right side $- 2$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

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

The left side $3$ is greater than the right side $2$ , which means that the given statement is false.

False

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

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

$x = 4$

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

$\frac{{(4)}^{2} + 2 (4) + 3}{(4) + 1} \leq (4) + 2$

The left side $5.4$ is less than the right side $6$ , which means that the given statement is always true.

True

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

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or x \geq 1$

Interval Notation:

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

Solve the Inequality for x (x^2+2x+3)/(x+1)<=x+2

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