Solve the Inequality for x (-x)/(x-6)<=1/(x-2)

$\frac{- x}{x - 6} \leq \frac{1}{x - 2}$

Step 1

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

$\frac{- x}{x - 6} - \frac{1}{x - 2} \leq 0$

Step 2

Simplify $\frac{- x}{x - 6} - \frac{1}{x - 2}$ .

Move the negative in front of the fraction.

$- \frac{x}{x - 6} - \frac{1}{x - 2} \leq 0$

To write $- \frac{x}{x - 6}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$- \frac{x}{x - 6} \cdot \frac{x - 2}{x - 2} - \frac{1}{x - 2} \leq 0$

To write $- \frac{1}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{x - 6}{x - 6}$ .

$- \frac{x}{x - 6} \cdot \frac{x - 2}{x - 2} - \frac{1}{x - 2} \cdot \frac{x - 6}{x - 6} \leq 0$

Write each expression with a common denominator of $(x - 6) (x - 2)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x}{x - 6}$ by $\frac{x - 2}{x - 2}$ .

$- \frac{x (x - 2)}{(x - 6) (x - 2)} - \frac{1}{x - 2} \cdot \frac{x - 6}{x - 6} \leq 0$

Multiply $\frac{1}{x - 2}$ by $\frac{x - 6}{x - 6}$ .

$- \frac{x (x - 2)}{(x - 6) (x - 2)} - \frac{x - 6}{(x - 2) (x - 6)} \leq 0$

Reorder the factors of $(x - 6) (x - 2)$ .

$- \frac{x (x - 2)}{(x - 2) (x - 6)} - \frac{x - 6}{(x - 2) (x - 6)} \leq 0$

Combine the numerators over the common denominator.

$\frac{- x (x - 2) - (x - 6)}{(x - 2) (x - 6)} \leq 0$

Simplify the numerator.

Apply the distributive property.

$\frac{- x \cdot x - x \cdot - 2 - (x - 6)}{(x - 2) (x - 6)} \leq 0$

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

Move $x$ .

$\frac{- (x \cdot x) - x \cdot - 2 - (x - 6)}{(x - 2) (x - 6)} \leq 0$

Multiply $x$ by $x$ .

$\frac{- x^{2} - x \cdot - 2 - (x - 6)}{(x - 2) (x - 6)} \leq 0$

Multiply $- 2$ by $- 1$ .

$\frac{- x^{2} + 2 x - (x - 6)}{(x - 2) (x - 6)} \leq 0$

Apply the distributive property.

$\frac{- x^{2} + 2 x - x - - 6}{(x - 2) (x - 6)} \leq 0$

Multiply $- 1$ by $- 6$ .

$\frac{- x^{2} + 2 x - x + 6}{(x - 2) (x - 6)} \leq 0$

Subtract $x$ from $2 x$ .

$\frac{- x^{2} + x + 6}{(x - 2) (x - 6)} \leq 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 6 = - 6$ and whose sum is $b = 1$ .

Multiply by $1$ .

$\frac{- x^{2} + 1 x + 6}{(x - 2) (x - 6)} \leq 0$

Rewrite $1$ as $- 2$ plus $3$

$\frac{- x^{2} + (- 2 + 3) x + 6}{(x - 2) (x - 6)} \leq 0$

Apply the distributive property.

$\frac{- x^{2} - 2 x + 3 x + 6}{(x - 2) (x - 6)} \leq 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$\frac{(- x^{2} - 2 x) + 3 x + 6}{(x - 2) (x - 6)} \leq 0$

Factor out the greatest common factor (GCF) from each group.

$\frac{x (- x - 2) - 3 (- x - 2)}{(x - 2) (x - 6)} \leq 0$

Factor the polynomial by factoring out the greatest common factor, $- x - 2$ .

$\frac{(- x - 2) (x - 3)}{(x - 2) (x - 6)} \leq 0$

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

$\frac{(- (x) - 2) (x - 3)}{(x - 2) (x - 6)} \leq 0$

Rewrite $- 2$ as $- 1 (2)$ .

$\frac{(- (x) - 1 (2)) (x - 3)}{(x - 2) (x - 6)} \leq 0$

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

$\frac{- (x + 2) (x - 3)}{(x - 2) (x - 6)} \leq 0$

Rewrite $- (x + 2)$ as $- 1 (x + 2)$ .

$\frac{- 1 (x + 2) (x - 3)}{(x - 2) (x - 6)} \leq 0$

Move the negative in front of the fraction.

$- \frac{(x + 2) (x - 3)}{(x - 2) (x - 6)} \leq 0$

Step 3

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

$x + 2 = 0$

$x - 3 = 0$

$x - 2 = 0$

$x - 6 = 0$

Step 4

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 5

Add $3$ to both sides of the equation.

$x = 3$

Step 6

Add $2$ to both sides of the equation.

$x = 2$

Step 7

Add $6$ to both sides of the equation.

$x = 6$

Step 8

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

$x = - 2$

$x = 3$

$x = 2$

$x = 6$

Step 9

Consolidate the solutions.

$x = - 2, 3, 2, 6$

Step 10

Find the domain of $- \frac{(x + 2) (x - 3)}{(x - 2) (x - 6)}$ .

Set the denominator in $\frac{(x + 2) (x - 3)}{(x - 2) (x - 6)}$ equal to $0$ to find where the expression is undefined.

$(x - 2) (x - 6) = 0$

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 $0$ .

$x - 2 = 0$

$x - 6 = 0$

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.

$x = 2$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

The final solution is all the values that make $(x - 2) (x - 6) = 0$ true.

$x = 2, 6$

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

$(- \infty, 2) \cup (2, 6) \cup (6, \infty)$

Step 11

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 2$

$2 < x < 3$

$3 < x < 6$

$x > 6$

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 $x < - 2$ to see if it makes the inequality true.

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

$x = - 4$

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

$\frac{- (- 4)}{(- 4) - 6} \leq \frac{1}{(- 4) - 2}$

The left side $- 0.4$ is less than the right side $- 0.1\overline{6}$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$\frac{- (0)}{(0) - 6} \leq \frac{1}{(0) - 2}$

The left side $0$ is greater than the right side $- 0.5$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Replace $x$ with $2.5$ in the original inequality.

$\frac{- (2.5)}{(2.5) - 6} \leq \frac{1}{(2.5) - 2}$

The left side $0.\overline{714285}$ is less than the right side $2$ , which means that the given statement is always true.

True

Test a value on the interval $3 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 6$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{- (4)}{(4) - 6} \leq \frac{1}{(4) - 2}$

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

False

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

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

$\frac{- (8)}{(8) - 6} \leq \frac{1}{(8) - 2}$

The left side $- 4$ is less 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.

$x < - 2$ True

$- 2 < x < 2$ False

$2 < x < 3$ True

$3 < x < 6$ False

$x > 6$ True

$x < - 2$ True

$- 2 < x < 2$ False

$2 < x < 3$ True

$3 < x < 6$ False

$x > 6$ True

Step 13

The solution consists of all of the true intervals.

$x \leq - 2$ or $2 < x \leq 3$ or $x > 6$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or 2 < x \leq 3 or x > 6$

Interval Notation:

$(- \infty, - 2] \cup (2, 3] \cup (6, \infty)$

Step 15

Solve the Inequality for x (-x)/(x-6)<=1/(x-2)

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