Precalculus Examples

$x + \frac{12}{x} < 7$

Step 1

Find the LCD of the terms in the equation.

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Step 1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x, 1$

Step 1.2

The LCM of one and any expression is the expression.

$x$

Step 2

Multiply each term in $x + \frac{12}{x} < 7$ by $x$ to eliminate the fractions.

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Step 2.1

Multiply each term in $x + \frac{12}{x} < 7$ by $x$ .

$x \cdot x + \frac{12}{x} x < 7 x$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Multiply $x$ by $x$ .

$x^{2} + \frac{12}{x} x < 7 x$

Step 2.2.1.2

Cancel the common factor of $x$ .

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Step 2.2.1.2.1

Cancel the common factor.

$x^{2} + \frac{12}{x} x < 7 x$

Step 2.2.1.2.2

Rewrite the expression.

$x^{2} + 12 < 7 x$

Step 3

Solve the inequality.

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Step 3.1

Subtract $7 x$ from both sides of the inequality.

$x^{2} + 12 - 7 x < 0$

Step 3.2

Convert the inequality to an equation.

$x^{2} + 12 - 7 x = 0$

Step 3.3

Factor $x^{2} + 12 - 7 x$ using the AC method.

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Step 3.3.1

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 $12$ and whose sum is $- 7$ .

$- 4, - 3$

Step 3.3.2

Write the factored form using these integers.

$(x - 4) (x - 3) = 0$

Step 3.4

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

$x - 4 = 0$

$x - 3 = 0$

Step 3.5

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 3.5.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.6

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

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Step 3.6.1

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

$x - 3 = 0$

Step 3.6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 3.7

The final solution is all the values that make $(x - 4) (x - 3) = 0$ true.

$x = 4, 3$

Step 4

Find the domain of $x + \frac{12}{x} - 7$ .

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Step 4.1

Set the denominator in $\frac{12}{x}$ 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.

$x < 0$

$0 < x < 3$

$3 < x < 4$

$x > 4$

Step 6

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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Step 6.1

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

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Step 6.1.1

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

$x = - 2$

Step 6.1.2

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

$(- 2) + \frac{12}{- 2} < 7$

Step 6.1.3

The left side $- 8$ is less than the right side $7$ , which means that the given statement is always true.

True

Step 6.2

Test a value on the interval $0 < x < 3$ to see if it makes the inequality true.

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Step 6.2.1

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

$x = 2$

Step 6.2.2

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

$(2) + \frac{12}{2} < 7$

Step 6.2.3

The left side $8$ is not less than the right side $7$ , which means that the given statement is false.

False

Step 6.3

Test a value on the interval $3 < x < 4$ to see if it makes the inequality true.

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Step 6.3.1

Choose a value on the interval $3 < x < 4$ and see if this value makes the original inequality true.

$x = 3.5$

Step 6.3.2

Replace $x$ with $3.5$ in the original inequality.

$(3.5) + \frac{12}{3.5} < 7$

Step 6.3.3

The left side $6.9\overline{285714}$ is less than the right side $7$ , which means that the given statement is always true.

True

Step 6.4

Test a value on the interval $x > 4$ to see if it makes the inequality true.

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Step 6.4.1

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Step 6.4.2

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

$(6) + \frac{12}{6} < 7$

Step 6.4.3

The left side $8$ is not less than the right side $7$ , which means that the given statement is false.

False

Step 6.5

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

$x < 0$ True

$0 < x < 3$ False

$3 < x < 4$ True

$x > 4$ False

$x < 0$ True

$0 < x < 3$ False

$3 < x < 4$ True

$x > 4$ False

Step 7

The solution consists of all of the true intervals.

$x < 0$ or $3 < x < 4$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x < 0 or 3 < x < 4$

Interval Notation:

$(- \infty, 0) \cup (3, 4)$

Step 9