Precalculus Examples

$6 x - 5 < \frac{6}{x}$

Step 1

Multiply both sides by $x$ .

$(6 x - 5) x = \frac{6}{x} x$

Step 2

Simplify.

Tap for more steps...

Step 2.1

Simplify the left side.

Tap for more steps...

Step 2.1.1

Simplify $(6 x - 5) x$ .

Tap for more steps...

Step 2.1.1.1

Apply the distributive property.

$6 x \cdot x - 5 x = \frac{6}{x} x$

Step 2.1.1.2

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

Tap for more steps...

Step 2.1.1.2.1

Move $x$ .

$6 (x \cdot x) - 5 x = \frac{6}{x} x$

Step 2.1.1.2.2

Multiply $x$ by $x$ .

$6 x^{2} - 5 x = \frac{6}{x} x$

Step 2.2

Simplify the right side.

Tap for more steps...

Step 2.2.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 2.2.1.1

Cancel the common factor.

$6 x^{2} - 5 x = \frac{6}{x} x$

Step 2.2.1.2

Rewrite the expression.

$6 x^{2} - 5 x = 6$

Step 3

Solve for $x$ .

Tap for more steps...

Step 3.1

Subtract $6$ from both sides of the equation.

$6 x^{2} - 5 x - 6 = 0$

Step 3.2

Factor by grouping.

Tap for more steps...

Step 3.2.1

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 $a \cdot c = 6 \cdot - 6 = - 36$ and whose sum is $b = - 5$ .

Tap for more steps...

Step 3.2.1.1

Factor $- 5$ out of $- 5 x$ .

$6 x^{2} - 5 x - 6 = 0$

Step 3.2.1.2

Rewrite $- 5$ as $4$ plus $- 9$

$6 x^{2} + (4 - 9) x - 6 = 0$

Step 3.2.1.3

Apply the distributive property.

$6 x^{2} + 4 x - 9 x - 6 = 0$

Step 3.2.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 3.2.2.1

Group the first two terms and the last two terms.

$(6 x^{2} + 4 x) - 9 x - 6 = 0$

Step 3.2.2.2

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

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

Step 3.2.3

Factor the polynomial by factoring out the greatest common factor, $3 x + 2$ .

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

Step 3.3

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

$3 x + 2 = 0$

$2 x - 3 = 0$

Step 3.4

Set $3 x + 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 3.4.1

Set $3 x + 2$ equal to $0$ .

$3 x + 2 = 0$

Step 3.4.2

Solve $3 x + 2 = 0$ for $x$ .

Tap for more steps...

Step 3.4.2.1

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Step 3.4.2.2

Divide each term in $3 x = - 2$ by $3$ and simplify.

Tap for more steps...

Step 3.4.2.2.1

Divide each term in $3 x = - 2$ by $3$ .

$\frac{3 x}{3} = \frac{- 2}{3}$

Step 3.4.2.2.2

Simplify the left side.

Tap for more steps...

Step 3.4.2.2.2.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 3.4.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 2}{3}$

Step 3.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 2}{3}$

Step 3.4.2.2.3

Simplify the right side.

Tap for more steps...

Step 3.4.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

Step 3.5

Set $2 x - 3$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 3.5.1

Set $2 x - 3$ equal to $0$ .

$2 x - 3 = 0$

Step 3.5.2

Solve $2 x - 3 = 0$ for $x$ .

Tap for more steps...

Step 3.5.2.1

Add $3$ to both sides of the equation.

$2 x = 3$

Step 3.5.2.2

Divide each term in $2 x = 3$ by $2$ and simplify.

Tap for more steps...

Step 3.5.2.2.1

Divide each term in $2 x = 3$ by $2$ .

$\frac{2 x}{2} = \frac{3}{2}$

Step 3.5.2.2.2

Simplify the left side.

Tap for more steps...

Step 3.5.2.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{3}{2}$

Step 3.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{3}{2}$

Step 3.6

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

$x = - \frac{2}{3}, \frac{3}{2}$

Step 4

Find the domain of $6 x - 5 - \frac{6}{x}$ .

Tap for more steps...

Step 4.1

Set the denominator in $\frac{6}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

Step 4.2

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

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

Step 5

Use each root to create test intervals.

$x < - \frac{2}{3}$

$- \frac{2}{3} < x < 0$

$0 < x < \frac{3}{2}$

$x > \frac{3}{2}$

Step 6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 6.1

Test a value on the interval $x < - \frac{2}{3}$ to see if it makes the inequality true.

Tap for more steps...

Step 6.1.1

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

$x = - 3$

Step 6.1.2

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

$6 (- 3) - 5 < \frac{6}{- 3}$

Step 6.1.3

The left side $- 23$ is less than the right side $- 2$ , which means that the given statement is always true.

True

Step 6.2

Test a value on the interval $- \frac{2}{3} < x < 0$ to see if it makes the inequality true.

Tap for more steps...

Step 6.2.1

Choose a value on the interval $- \frac{2}{3} < x < 0$ and see if this value makes the original inequality true.

$x = - 0.33$

Step 6.2.2

Replace $x$ with $- 0.33$ in the original inequality.

$6 (- 0.33) - 5 < \frac{6}{- 0.33}$

Step 6.2.3

The left side $- 6.98$ is not less than the right side $- 18.\overline{18}$ , which means that the given statement is false.

False

Step 6.3

Test a value on the interval $0 < x < \frac{3}{2}$ to see if it makes the inequality true.

Tap for more steps...

Step 6.3.1

Choose a value on the interval $0 < x < \frac{3}{2}$ and see if this value makes the original inequality true.

$x = 1$

Step 6.3.2

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

$6 (1) - 5 < \frac{6}{1}$

Step 6.3.3

The left side $1$ is less than the right side $6$ , which means that the given statement is always true.

True

Step 6.4

Test a value on the interval $x > \frac{3}{2}$ to see if it makes the inequality true.

Tap for more steps...

Step 6.4.1

Choose a value on the interval $x > \frac{3}{2}$ and see if this value makes the original inequality true.

$x = 4$

Step 6.4.2

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

$6 (4) - 5 < \frac{6}{4}$

Step 6.4.3

The left side $19$ is not less than the right side $1.5$ , which means that the given statement is false.

False

Step 6.5

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

$x < - \frac{2}{3}$ True

$- \frac{2}{3} < x < 0$ False

$0 < x < \frac{3}{2}$ True

$x > \frac{3}{2}$ False

$x < - \frac{2}{3}$ True

$- \frac{2}{3} < x < 0$ False

$0 < x < \frac{3}{2}$ True

$x > \frac{3}{2}$ False

Step 7

The solution consists of all of the true intervals.

$x < - \frac{2}{3}$ or $0 < x < \frac{3}{2}$

Step 8

Convert the inequality to interval notation.

$(- \infty, - \frac{2}{3}) \cup (0, \frac{3}{2})$

Step 9