Convert to Interval Notation 1/(-x)

$\frac{1}{- x} \leq \frac{5}{7 - x}$

Move $\frac{5}{7 - x}$ to the left side of the equation by subtracting it from both sides.

$\frac{1}{- x} - \frac{5}{7 - x} \leq 0$

Simplify $\frac{1}{- x} - \frac{5}{7 - x}$ .

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

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

$\frac{- 1 (- 1)}{- x} - \frac{5}{7 - x} \leq 0$

Move the negative in front of the fraction.

$- \frac{1}{x} - \frac{5}{7 - x} \leq 0$

To write $- \frac{1}{x}$ as a fraction with a common denominator, multiply by $\frac{7 - x}{7 - x}$ .

$- \frac{1}{x} \cdot \frac{7 - x}{7 - x} - \frac{5}{7 - x} \leq 0$

To write $- \frac{5}{7 - x}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$- \frac{1}{x} \cdot \frac{7 - x}{7 - x} - \frac{5}{7 - x} \cdot \frac{x}{x} \leq 0$

Write each expression with a common denominator of $x (7 - x)$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{1}{x}$ and $\frac{7 - x}{7 - x}$ .

$- \frac{7 - x}{x (7 - x)} - \frac{5}{7 - x} \cdot \frac{x}{x} \leq 0$

Multiply $\frac{5}{7 - x}$ and $\frac{x}{x}$ .

$- \frac{7 - x}{x (7 - x)} - \frac{5 x}{(7 - x) x} \leq 0$

Reorder the factors of $(7 - x) x$ .

$- \frac{7 - x}{x (7 - x)} - \frac{5 x}{x (7 - x)} \leq 0$

Combine the numerators over the common denominator.

$\frac{- (7 - x) - 5 x}{x (7 - x)} \leq 0$

Simplify the numerator.

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Apply the distributive property.

$\frac{- 1 \cdot 7 - - x - 5 x}{x (7 - x)} \leq 0$

Multiply $- 1$ by $7$ .

$\frac{- 7 - - x - 5 x}{x (7 - x)} \leq 0$

Multiply $- - x$ .

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Multiply $- 1$ by $- 1$ .

$\frac{- 7 + 1 x - 5 x}{x (7 - x)} \leq 0$

Multiply $x$ by $1$ .

$\frac{- 7 + x - 5 x}{x (7 - x)} \leq 0$

Subtract $5 x$ from $x$ .

$\frac{- 7 - 4 x}{x (7 - x)} \leq 0$

Simplify with factoring out.

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Rewrite $- 7$ as $- 1 (7)$ .

$\frac{- 1 (7) - 4 x}{x (7 - x)} \leq 0$

Factor $- 1$ out of $- 4 x$ .

$\frac{- 1 (7) - (4 x)}{x (7 - x)} \leq 0$

Factor $- 1$ out of $- 1 (7) - (4 x)$ .

$\frac{- 1 (7 + 4 x)}{x (7 - x)} \leq 0$

Move the negative in front of the fraction.

$- \frac{7 + 4 x}{x (7 - x)} \leq 0$

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

$x = 0$

$7 + 4 x = 0$

$7 - x = 0$

Subtract $7$ from both sides of the equation.

$4 x = - 7$

Divide each term by $4$ and simplify.

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Divide each term in $4 x = - 7$ by $4$ .

$\frac{4 x}{4} = \frac{- 7}{4}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 x}{4} = \frac{- 7}{4}$

Divide $x$ by $1$ .

$x = \frac{- 7}{4}$

Move the negative in front of the fraction.

$x = - \frac{7}{4}$

Subtract $7$ from both sides of the equation.

$- x = - 7$

Multiply each term in $- x = - 7$ by $- 1$

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Multiply each term in $- x = - 7$ by $- 1$ .

$(- x) \cdot - 1 = (- 7) \cdot - 1$

Multiply $(- x) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 x = (- 7) \cdot - 1$

Multiply $x$ by $1$ .

$x = (- 7) \cdot - 1$

Multiply $- 7$ by $- 1$ .

$x = 7$

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

$x = 0$

$x = - \frac{7}{4}$

$x = 7$

Consolidate the solutions.

$x = 0, - \frac{7}{4}, 7$

Find the domain of $- \frac{7 + 4 x}{x (7 - x)}$ .

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Set the denominator in $\frac{7 + 4 x}{x (7 - x)}$ equal to $0$ to find where the expression is undefined.

$x (7 - x) = 0$

Solve for $x$ .

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If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$7 - x = 0$

Set the first factor equal to $0$ .

$x = 0$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$7 - x = 0$

Subtract $7$ from both sides of the equation.

$- x = - 7$

Multiply each term in $- x = - 7$ by $- 1$

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Multiply each term in $- x = - 7$ by $- 1$ .

$(- x) \cdot - 1 = (- 7) \cdot - 1$

Multiply $(- x) \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 x = (- 7) \cdot - 1$

Multiply $x$ by $1$ .

$x = (- 7) \cdot - 1$

Multiply $- 7$ by $- 1$ .

$x = 7$

The final solution is all the values that make $x (7 - x) = 0$ true.

$x = 0, 7$

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

$(- \infty, 0) \cup (0, 7) \cup (7, \infty)$

Use each root to create test intervals.

$x < - \frac{7}{4}$

$- \frac{7}{4} < x < 0$

$0 < x < 7$

$x > 7$

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 $x < - \frac{7}{4}$ to see if it makes the inequality true.

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Choose a value on the interval $x < - \frac{7}{4}$ and see if this value makes the original inequality true.

$x = - 4$

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

$\frac{1}{- (- 4)} \leq \frac{5}{7 - (- 4)}$

The left side $0.25$ is less than the right side $0.\overline{45}$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{7}{4} < x < 0$ to see if it makes the inequality true.

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Choose a value on the interval $- \frac{7}{4} < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

Replace $x$ with $- 1$ in the original inequality.

$\frac{1}{- (- 1)} \leq \frac{5}{7 - (- 1)}$

The left side $1$ is greater than the right side $0.625$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 7$ to see if it makes the inequality true.

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Choose a value on the interval $0 < x < 7$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{1}{- (4)} \leq \frac{5}{7 - (4)}$

The left side $- 0.25$ is less than the right side $1.\overline{6}$ , which means that the given statement is always true.

True

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$\frac{1}{- (10)} \leq \frac{5}{7 - (10)}$

The left side $- 0.1$ is greater than the right side $- 1.\overline{6}$ , which means that the given statement is false.

False

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

$x < - \frac{7}{4}$ True

$- \frac{7}{4} < x < 0$ False

$0 < x < 7$ True

$x > 7$ False

$x < - \frac{7}{4}$ True

$- \frac{7}{4} < x < 0$ False

$0 < x < 7$ True

$x > 7$ False

The solution consists of all of the true intervals.

$x \leq - \frac{7}{4}$ or $0 < x < 7$

Convert the inequality to interval notation.

$(- \infty, - \frac{7}{4}] \cup (0, 7)$

Convert to Interval Notation 1/(-x)<=5/(7-x)

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