Solve the Inequality for x -2/(5-x)>=1/(x-2)

$\frac{- 2}{5 - x} \geq \frac{1}{x - 2}$

Step 1

Subtract $\frac{1}{x - 2}$ from both sides of the inequality.

$\frac{- 2}{5 - x} - \frac{1}{x - 2} \geq 0$

Step 2

Simplify $\frac{- 2}{5 - x} - \frac{1}{x - 2}$ .

Move the negative in front of the fraction.

$- \frac{2}{5 - x} - \frac{1}{x - 2} \geq 0$

To write $- \frac{2}{5 - x}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$- \frac{2}{5 - x} \cdot \frac{x - 2}{x - 2} - \frac{1}{x - 2} \geq 0$

To write $- \frac{1}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{5 - x}{5 - x}$ .

$- \frac{2}{5 - x} \cdot \frac{x - 2}{x - 2} - \frac{1}{x - 2} \cdot \frac{5 - x}{5 - x} \geq 0$

Write each expression with a common denominator of $(5 - x) (x - 2)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2}{5 - x}$ by $\frac{x - 2}{x - 2}$ .

$- \frac{2 (x - 2)}{(5 - x) (x - 2)} - \frac{1}{x - 2} \cdot \frac{5 - x}{5 - x} \geq 0$

Multiply $\frac{1}{x - 2}$ by $\frac{5 - x}{5 - x}$ .

$- \frac{2 (x - 2)}{(5 - x) (x - 2)} - \frac{5 - x}{(x - 2) (5 - x)} \geq 0$

Reorder the factors of $(x - 2) (5 - x)$ .

$- \frac{2 (x - 2)}{(5 - x) (x - 2)} - \frac{5 - x}{(5 - x) (x - 2)} \geq 0$

Combine the numerators over the common denominator.

$\frac{- 2 (x - 2) - (5 - x)}{(5 - x) (x - 2)} \geq 0$

Simplify the numerator.

Factor $- 1$ out of $- 2 (x - 2) - (5 - x)$ .

Reorder $5$ and $- x$ .

$\frac{- 2 (x - 2) - (- x + 5)}{(5 - x) (x - 2)} \geq 0$

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

$\frac{- (2 (x - 2)) - (- x + 5)}{(5 - x) (x - 2)} \geq 0$

Factor $- 1$ out of $- (2 (x - 2)) - (- x + 5)$ .

$\frac{- (2 (x - 2) - x + 5)}{(5 - x) (x - 2)} \geq 0$

Apply the distributive property.

$\frac{- (2 x + 2 \cdot - 2 - x + 5)}{(5 - x) (x - 2)} \geq 0$

Multiply $2$ by $- 2$ .

$\frac{- (2 x - 4 - x + 5)}{(5 - x) (x - 2)} \geq 0$

Subtract $x$ from $2 x$ .

$\frac{- (x - 4 + 5)}{(5 - x) (x - 2)} \geq 0$

Add $- 4$ and $5$ .

$\frac{- (x + 1)}{(5 - x) (x - 2)} \geq 0$

Move the negative in front of the fraction.

$- \frac{x + 1}{(5 - x) (x - 2)} \geq 0$

Step 3

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

$x + 1 = 0$

$5 - x = 0$

$x - 2 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

Subtract $5$ from both sides of the equation.

$- x = - 5$

Step 6

Divide each term in $- x = - 5$ by $- 1$ and simplify.

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Simplify the right side.

Divide $- 5$ by $- 1$ .

$x = 5$

Step 7

Add $2$ to both sides of the equation.

$x = 2$

Step 8

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

$x = - 1$

$x = 5$

$x = 2$

Step 9

Consolidate the solutions.

$x = - 1, 5, 2$

Step 10

Find the domain of $- \frac{x + 1}{(5 - x) (x - 2)}$ .

Set the denominator in $\frac{x + 1}{(5 - x) (x - 2)}$ equal to $0$ to find where the expression is undefined.

$(5 - x) (x - 2) = 0$

Solve for $x$ .

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

$5 - x = 0$

$x - 2 = 0$

Set $5 - x$ equal to $0$ and solve for $x$ .

Set $5 - x$ equal to $0$ .

$5 - x = 0$

Solve $5 - x = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$- x = - 5$

Divide each term in $- x = - 5$ by $- 1$ and simplify.

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Simplify the right side.

Divide $- 5$ by $- 1$ .

$x = 5$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(5 - x) (x - 2) = 0$ true.

$x = 5, 2$

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

$(- \infty, 2) \cup (2, 5) \cup (5, \infty)$

Step 11

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 2$

$2 < x < 5$

$x > 5$

Step 12

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.

$\frac{- 2}{5 - (- 4)} \geq \frac{1}{(- 4) - 2}$

The left side $- 0.\overline{2}$ is less than the right side $- 0.1\overline{6}$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{- 2}{5 - (0)} \geq \frac{1}{(0) - 2}$

The left side $- 0.4$ is greater than the right side $- 0.5$ , which means that the given statement is always true.

True

Test a value on the interval $2 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{- 2}{5 - (4)} \geq \frac{1}{(4) - 2}$

The left side $- 2$ is less than the right side $0.5$ , which means that the given statement is false.

False

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

$\frac{- 2}{5 - (8)} \geq \frac{1}{(8) - 2}$

The left side $0.\overline{6}$ is greater than the right side $0.1\overline{6}$ , which means that the given statement is always true.

True

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

$x < - 1$ False

$- 1 < x < 2$ True

$2 < x < 5$ False

$x > 5$ True

$x < - 1$ False

$- 1 < x < 2$ True

$2 < x < 5$ False

$x > 5$ True

Step 13

The solution consists of all of the true intervals.

$- 1 \leq x < 2$ or $x > 5$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$- 1 \leq x < 2 or x > 5$

Interval Notation:

$[- 1, 2) \cup (5, \infty)$

Step 15

Solve the Inequality for x -2/(5-x)>=1/(x-2)

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