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

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

Step 1

Multiply both sides by $x^{2} - x$ .

$\frac{1}{x^{2} - x} (x^{2} - x) = \frac{1}{6} (x^{2} - x)$

Step 2

Simplify.

Simplify the left side.

Cancel the common factor of $x^{2} - x$ .

Cancel the common factor.

$\frac{1}{x^{2} - x} (x^{2} - x) = \frac{1}{6} (x^{2} - x)$

Rewrite the expression.

$1 = \frac{1}{6} (x^{2} - x)$

Simplify the right side.

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

Apply the distributive property.

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

Combine $\frac{1}{6}$ and $x^{2}$ .

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

Combine $\frac{1}{6}$ and $x$ .

$1 = \frac{x^{2}}{6} - \frac{x}{6}$

Step 3

Solve for $x$ .

Rewrite the equation as $\frac{x^{2}}{6} - \frac{x}{6} = 1$ .

$\frac{x^{2}}{6} - \frac{x}{6} = 1$

Multiply each term in $\frac{x^{2}}{6} - \frac{x}{6} = 1$ by $6$ to eliminate the fractions.

Multiply each term in $\frac{x^{2}}{6} - \frac{x}{6} = 1$ by $6$ .

$\frac{x^{2}}{6} \cdot 6 - \frac{x}{6} \cdot 6 = 1 \cdot 6$

Simplify the left side.

Simplify each term.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{x^{2}}{6} \cdot 6 - \frac{x}{6} \cdot 6 = 1 \cdot 6$

Rewrite the expression.

$x^{2} - \frac{x}{6} \cdot 6 = 1 \cdot 6$

Cancel the common factor of $6$ .

Move the leading negative in $- \frac{x}{6}$ into the numerator.

$x^{2} + \frac{- x}{6} \cdot 6 = 1 \cdot 6$

Cancel the common factor.

$x^{2} + \frac{- x}{6} \cdot 6 = 1 \cdot 6$

Rewrite the expression.

$x^{2} - x = 1 \cdot 6$

Simplify the right side.

Multiply $6$ by $1$ .

$x^{2} - x = 6$

Subtract $6$ from both sides of the equation.

$x^{2} - x - 6 = 0$

Factor $x^{2} - x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 1$ .

$- 3, 2$

Write the factored form using these integers.

$(x - 3) (x + 2) = 0$

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

$x - 3 = 0$

$x + 2 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $(x - 3) (x + 2) = 0$ true.

$x = 3, - 2$

Step 4

Find the domain of $\frac{1}{x (x - 1)} - \frac{1}{6}$ .

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

$x (x - 1) = 0$

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 $0$ .

$x = 0$

$x - 1 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $x (x - 1) = 0$ true.

$x = 0, 1$

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

$(- \infty, 0) \cup (0, 1) \cup (1, \infty)$

Step 5

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 0$

$0 < x < 1$

$1 < x < 3$

$x > 3$

Step 6

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.

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

The left side $0.05$ 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 $- 2 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 2 < x < 0$ and see if this value makes the original inequality true.

$x = - 1$

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

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

The left side $0.5$ is greater than the right side $0.1\overline{6}$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Replace $x$ with $0.5$ in the original inequality.

$\frac{1}{{(0.5)}^{2} - (0.5)} \leq \frac{1}{6}$

The left side $- 4$ 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 $1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

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

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

The left side $0.5$ is greater than the right side $0.1\overline{6}$ , which means that the given statement is false.

False

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

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

$x = 6$

Replace $x$ with $6$ in the original inequality.

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

The left side $0.0\overline{3}$ 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

$- 2 < x < 0$ False

$0 < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

$x < - 2$ True

$- 2 < x < 0$ False

$0 < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

Step 7

The solution consists of all of the true intervals.

$x \leq - 2$ or $0 < x < 1$ or $x \geq 3$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 2 or 0 < x < 1 or x \geq 3$

Interval Notation:

$(- \infty, - 2] \cup (0, 1) \cup [3, \infty)$

Step 9

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

Related Questions

Support LCMGCF.com by disabling your adblocker.