Inequalities -3/(-3x)-3/3>=-3

$\frac{- 3}{- 3 x} - \frac{3}{3} \geq - 3$

Simplify each term.

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Cancel the common factor of $3$ and $- 3$ .

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Factor $3$ out of $3$ .

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

Cancel the common factors.

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Factor $3$ out of $- 3 x$ .

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

Cancel the common factor.

$- \frac{3 \cdot 1}{3 (- x)} - \frac{3}{3} \geq - 3$

Rewrite the expression.

$- \frac{1}{- x} - \frac{3}{3} \geq - 3$

Cancel the common factor of $1$ and $- 1$ .

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

$- \frac{- 1 \cdot - 1}{- x} - \frac{3}{3} \geq - 3$

Move the negative in front of the fraction.

$\frac{1}{x} - \frac{3}{3} \geq - 3$

Multiply $- - \frac{1}{x}$ .

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Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{1}{x}$ by $1$ .

$\frac{1}{x} - \frac{3}{3} \geq - 3$

Divide $3$ by $3$ .

$\frac{1}{x} - 1 \cdot 1 \geq - 3$

Multiply $- 1$ by $1$ .

$\frac{1}{x} - 1 \geq - 3$

Move all terms not containing $x$ to the right side of the inequality.

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Add $1$ to both sides of the inequality.

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

Add $- 3$ and $1$ .

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

Solve for $x$ .

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Multiply each term by $x$ and simplify.

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Multiply each term in $\frac{1}{x} \geq - 2$ by $x$ .

$\frac{1}{x} \cdot x \geq - 2 \cdot x$

Cancel the common factor of $x$ .

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Cancel the common factor.

$\frac{1}{x} \cdot x \geq - 2 \cdot x$

Rewrite the expression.

$1 \geq - 2 \cdot x$

$1 \geq - 2 x$

Rewrite so $x$ is on the left side of the inequality.

$- 2 x \leq 1$

Divide each term by $- 2$ and simplify.

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Divide each term in $- 2 x \leq 1$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Cancel the common factor of $- 2$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

Move the negative in front of the fraction.

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

Find the domain of $\frac{1}{x} + 2$ .

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

$x = 0$

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

Interval Notation:

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

Interval Notation:

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

Use each root to create test intervals.

$x \leq - \frac{1}{2}$

$- \frac{1}{2} \leq x < 0$

$x > 0$

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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Test a value on the interval $x \leq - \frac{1}{2}$ to see if it makes the inequality true.

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Choose a value on the interval $x \leq - \frac{1}{2}$ and see if this value makes the original inequality true.

$x = - 3$

Replace $x$ with $- 3$ in the original inequality.

$\frac{- 3}{- 3 \cdot - 3} - \frac{3}{3} \geq - 3$

The left side $- 1.\overline{3}$ is greater than the right side $- 3$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{1}{2} \leq x < 0$ to see if it makes the inequality true.

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Choose a value on the interval $- \frac{1}{2} \leq x < 0$ and see if this value makes the original inequality true.

$x = - 0.25$

Replace $x$ with $- 0.25$ in the original inequality.

$\frac{- 3}{- 3 \cdot - 0.25} - \frac{3}{3} \geq - 3$

The left side $- 5$ is less than the right side $- 3$ , which means that the given statement is false.

False

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

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

$x = 2$

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

$\frac{- 3}{- 3 \cdot 2} - \frac{3}{3} \geq - 3$

The left side $- 0.5$ is greater than the right side $- 3$ , which means that the given statement is always true.

True

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

$x \leq - \frac{1}{2}$ True

$- \frac{1}{2} \leq x < 0$ False

$x > 0$ True

$x \leq - \frac{1}{2}$ True

$- \frac{1}{2} \leq x < 0$ False

$x > 0$ True

The solution consists of all of the true intervals.

$x \leq - \frac{1}{2}$ or $x > 0$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - \frac{1}{2} or x > 0$

Interval Notation:

$(- \infty, - \frac{1}{2}] \cup (0, \infty)$

Inequalities -3/(-3x)-3/3>=-3

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