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Solve Rational Inequalities

Solve rational inequalities

Solve Rational Inequalities

We learned to solve linear inequalities after learning to solve linear equations. The techniques were very much the same with one major exception. When we multiplied or divided by a negative number, the inequality sign reversed.

Having just learned to solve rational equations we are now ready to solve rational inequalities. A rational inequality is an inequality that contains a rational expression.

Inequalities such as $\frac{3}{2x}>1,\ \frac{2x}{x-3}<4,\ \frac{2x-3}{x-6}\ge x,$ and $\frac{1}{4}-\frac{2}{{x}^{2}}\le \frac{3}{x}$ are rational inequalities as they each contain a rational expression.

When we solve a rational inequality, we will use many of the techniques we used solving linear inequalities. We especially must remember that when we multiply or divide by a negative number, the inequality sign must reverse.

Another difference is that we must carefully consider what value might make the rational expression undefined and so must be excluded.

When we solve an equation and the result is $x=3,$ we know there is one solution, which is 3.

When we solve an inequality and the result is $x>3,$ we know there are many solutions. We graph the result to better help show all the solutions, and we start with 3. Three becomes a zero partition number and then we decide whether to shade to the left or right of it. The numbers to the right of 3 are larger than 3, so we shade to the right.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Solve an Inequality with Rational Functions

When working with rational functions, it is sometimes useful to know when the function is greater than or less than a particular value. This leads to a rational inequality.

Example

Try it.

Given the function $R(x)=\frac{x+3}{x-5},$ find the values of x that make the function less than or equal to 0.

Solution

We want the function to be less than or equal to 0.

	$R(x)\le 0$
Substitute the rational expression for $R(x).$	$\frac{x+3}{x-5}\le 0\ x\ne 5$
Find the zero partition numbers.	$\begin{array}{llllllll}x+3 & = & 0 & & & x-5 & = & 0 \\ x & = & -3 & & & x & = & 5\end{array}$
Use the zero partition numbers to divide the number line into intervals.

Test values in each interval. Above the number line, show the sign of each factor in each interval. Below the number line, show the sign of the quotient
Write the solution in interval notation. Since 5 is excluded we, do not include it in the interval.	$[-3,5)$

In economics, the function $C(x)$ is used to represent the cost of producing x units of a commodity. The average cost per unit can be found by dividing $C(x)$ by the number of items $x.$ Then, the average cost per unit is $c(x)=\frac{C(x)}{x}.$

Example

Try it.

The function $C(x)=10x+3000$ represents the cost to produce $x,$ number of items. Find ⓐ the average cost function, $c(x)$ ⓑ how many items should be produced so that the average cost is less than $40.

Solution

ⓐ

	$C(x)=10x+3000$
The average cost function is $c(x)=\frac{C(x)}{x}.$
To find the average cost function, divide the cost function by $x.$	$\begin{array}{l} \\ \\ c(x)=\frac{C(x)}{x} \\ c(x)=\frac{10x+3000}{x}\end{array}$
	The average cost function is $c(x)=\frac{10x+3000}{x}.$

ⓑ

We want the function $c(x)$ to be less than $40.$	$\ c(x)<40$
Substitute the rational expression for $c(x).$	$\ \frac{10x+3000}{x}<40\ x\ne 0$
Subtract 40 to get 0 on the right.	$\ \frac{10x+3000}{x}-40<0$
Rewrite the left side as one quotient by finding the LCD and performing the subtraction.	$\frac{10x+3000}{x}-40(\frac{x}{x})<0$
	$\ \frac{10x+3000}{x}-\frac{40x}{x}<0$
	$\ \frac{10x+3000-40x}{x}<0$
	$\ \frac{-30x+3000}{x}<0$
Factor the numerator to show all factors.	$\ \frac{-30(x-100)}{x}<0$
Find the zero partition numbers.	$\begin{array}{llll}-30(x-100) & = & 0 & \ x=0 \\ -30\ne 0\ x-100 & = & 0 & \\ x & = & 100 & \end{array}$

More than 100 items must be produced to keep the average cost below $40 per item.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Solve a rational inequality.
1. Write the inequality as one quotient on the left and zero on the right.
2. Determine the zero partition numbers–the points where the rational expression will be zero or undefined.
3. Use the zero partition numbers to divide the number line into intervals.
4. Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.
5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the value of $x-5$ when ⓐ $x=6$ ⓑ $x=-3$ ⓒ $x=5.$
If you missed this problem, review .

Otkrij odgovor

ⓐ 1; ⓑ −8; ⓒ 0
Solve: $8-2x<12.$
If you missed this problem, review .

Otkrij odgovor

$x>-2$
Write in interval notation: $-3\le x<5.$
If you missed this problem, review .

Otkrij odgovor

$[-3,\ 5)$
Solve and write the solution in interval notation: $\frac{x-1}{x+3}\ge 0.$

Otkrij odgovor

Step 1. Write the inequality as one quotient on the left and zero on the right.
Our inequality is in this form. $\ \frac{x-1}{x+3}\ge 0$
Step 2. Determine the zero partition numbers—the points where the rational expression will be zero or undefined.
The rational expression will be zero when the numerator is zero. Since $x-1=0$ when $x=1,$ then $1$ is a zero partition number.
The rational expression will be undefined when the denominator is zero. Since $x+3=0$ when $x=-3,$ then $-3$ is a zero partition number.
The zero partition numbers are 1 and $-3.$
Step 3. Use the zero partition numbers to divide the number line into intervals.
The number line is divided into three intervals:
$\ (\text{-}\infty ,-3)\ (-3,1)\ (1,\infty )$
Step 4. Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.
To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.
\[\text{Interval}\ (\text{-}\infty ,-3)\]
The number $-4$ is in the interval $(\text{-}\infty ,-3).$ Test $x=-4$ in the expression in the numerator and the denominator.
Above the number line, mark the factor $x-1$ negative and mark the factor $x+3$ negative.
Since a negative divided by a negative is positive, mark the quotient positive in the interval $(\text{-}\infty ,-3).$
\[\text{Interval}\ (-3,1)\]
The number 0 is in the interval $(-3,1).$ Test $x=0.$
Above the number line, mark the factor $x-1$ negative and mark $x+3$ positive.
Since a negative divided by a positive is negative, the quotient is marked negative in the interval $(-3,1).$
\[\text{Interval}\ (1,\infty )\]
The number 2 is in the interval $(1,\infty ).$ Test $x=2.$
Above the number line, mark the factor $x-1$ positive and mark $x+3$ positive.
Since a positive divided by a positive is positive, mark the quotient positive in the interval $(1,\infty ).$
Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.
We want the quotient to be greater than or equal to zero, so the numbers in the intervals $(\text{-}\infty ,-3)$ and $(1,\infty )$ are solutions.
But what about the zero partition numbers?
The zero partition number $x=-3$ makes the denominator 0, so it must be excluded from the solution and we mark it with a parenthesis.
The zero partition number $x=1$ makes the whole rational expression 0. The inequality requires that the rational expression be greater than or equal to 0. So, 1 is part of the solution and we will mark it with a bracket.
Recall that when we have a solution made up of more than one interval we use the union symbol, $\cup ,$ to connect the two intervals. The solution in interval notation is $(\text{-}\infty ,-3)\cup [1,\infty ).$
Solve and write the solution in interval notation: $\frac{x-2}{x+4}\ge 0.$

Otkrij odgovor

$(\text{-}\infty ,-4)\cup [2,\infty )$
Solve and write the solution in interval notation: $\frac{x+2}{x-4}\ge 0.$

Otkrij odgovor

$(\text{-}\infty ,-2]\cup (4,\infty )$

Solve and write the solution in interval notation: $\frac{4x}{x-6}<1.$

Otkrij odgovor

	$\frac{4x}{x-6}<1$
Subtract 1 to get zero on the right.	$\frac{4x}{x-6}-1<0$
Rewrite 1 as a fraction using the LCD.	$\frac{4x}{x-6}-\frac{x-6}{x-6}<0$
Subtract the numerators and place the difference over the common denominator.	$\frac{4x-(x-6)}{x-6}<0$
Simplify.	$\frac{3x+6}{x-6}<0$
Factor the numerator to show all factors.	$\frac{3(x+2)}{x-6}<0$
Find the zero partition numbers.
The quotient will be zero when the numerator is zero. The quotient is undefined when the denominator is zero.	$\begin{array}{llllllll}x+2 & = & 0 & & & x-6 & = & 0 \\ x & = & \text{-}2 & & & x & = & 6\end{array}$
Use the zero partition numbers to divide the number line into intervals.

Test a value in each interval.

Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

Determine the intervals where the inequality is correct. We want the quotient to be negative, so the solution includes the points between −2 and 6. Since the inequality is strictly less than, the endpoints are not included.
We write the solution in interval notation as (−2, 6).

Solve and write the solution in interval notation: $\frac{3x}{x-3}<1.$

Otkrij odgovor

$(-\frac{3}{2},3)$
Solve and write the solution in interval notation: $\frac{3x}{x-4}<2.$

Otkrij odgovor

$(-8,4)$

Solve and write the solution in interval notation: $\frac{5}{{x}^{2}-2x-15}>0.$

Otkrij odgovor

The inequality is in the correct form.	$\frac{5}{{x}^{2}-2x-15}>0$
Factor the denominator.	$\frac{5}{(x+3)(x-5)}>0$
Find the zero partition numbers. The quotient is 0 when the numerator is 0. Since the numerator is always 5, the quotient cannot be 0.
The quotient will be undefined when the denominator is zero.	$\begin{array}{l}(x+3)(x-5)=0 \\ x=-3,\ x=5\end{array}$
Use the zero partition numbers to divide the number line into intervals.

Test values in each interval. Above the number line show the sign of each factor of the denominator in each interval. Below the number line, show the sign of the quotient.
Write the solution in interval notation.	$(\text{-}\infty ,-3)\cup (5,\infty )$

Solve and write the solution in interval notation: $\frac{1}{{x}^{2}+2x-8}>0.$

Otkrij odgovor

$(\text{-}\infty ,-4)\cup (2,\infty )$
Solve and write the solution in interval notation: $\frac{3}{{x}^{2}+x-12}>0.$

Otkrij odgovor

$(\text{-}\infty ,-4)\cup (3,\infty )$

Solve and write the solution in interval notation: $\frac{1}{3}-\frac{2}{{x}^{2}}<\frac{5}{3x}.$

Otkrij odgovor

	$\frac{1}{3}-\frac{2}{{x}^{2}}<\frac{5}{3x}$
Subtract $\frac{5}{3x}$ to get zero on the right.	$\frac{1}{3}-\frac{2}{{x}^{2}}-\frac{5}{3x}<0$
Rewrite to get each fraction with the LCD $3{x}^{2}.$	$\frac{1⋅{x}^{2}}{3⋅{x}^{2}}-\frac{2⋅3}{{x}^{2}⋅3}-\frac{5⋅x}{3x⋅x}<0$
Simplify.	$\frac{{x}^{2}}{3{x}^{2}}-\frac{6}{3{x}^{2}}-\frac{5x}{3{x}^{2}}<0$
Subtract the numerators and place the difference over the common denominator.	$\frac{{x}^{2}-5x-6}{3{x}^{2}}<0$
Factor the numerator.	$\frac{(x-6)(x+1)}{3{x}^{2}}<0$
Find the zero partition numbers.	$\begin{array}{lllllllllllll} \\ 3{x}^{2} & = & 0 & & & x-6 & = & 0 & & & x+1 & = & 0 \\ x & = & 0 & & & x & = & 6 & & & x & = & \text{-}1\end{array}$
Use the zero partition numbers to divide the number line into intervals.

Above the number line show the sign of each factor in each interval. Below the number line, show the sign of the quotient.
Since, 0 is excluded, the solution is the two intervals, $(-1,0)$ and $(0,6).$	$(-1,0)\cup (0,6)$

Solve and write the solution in interval notation: $\frac{1}{2}+\frac{4}{{x}^{2}}<\frac{3}{x}.$

Otkrij odgovor

$(2,4)$
Solve and write the solution in interval notation: $\frac{1}{3}+\frac{6}{{x}^{2}}<\frac{3}{x}.$

Otkrij odgovor

$(3,6)$

Given the function $R(x)=\frac{x+3}{x-5},$ find the values of x that make the function less than or equal to 0.

Otkrij odgovor

We want the function to be less than or equal to 0.

	$R(x)\le 0$
Substitute the rational expression for $R(x).$	$\frac{x+3}{x-5}\le 0\ x\ne 5$
Find the zero partition numbers.	$\begin{array}{llllllll}x+3 & = & 0 & & & x-5 & = & 0 \\ x & = & -3 & & & x & = & 5\end{array}$
Use the zero partition numbers to divide the number line into intervals.

Test values in each interval. Above the number line, show the sign of each factor in each interval. Below the number line, show the sign of the quotient
Write the solution in interval notation. Since 5 is excluded we, do not include it in the interval.	$[-3,5)$

Given the function $R(x)=\frac{x-2}{x+4},$ find the values of x that make the function less than or equal to 0.

Otkrij odgovor

$(-4,2]$
Given the function $R(x)=\frac{x+1}{x-4},$ find the values of x that make the function less than or equal to 0.

Otkrij odgovor

$[-1,4)$

Otkrij odgovor

ⓐ

	$C(x)=10x+3000$
The average cost function is $c(x)=\frac{C(x)}{x}.$
To find the average cost function, divide the cost function by $x.$	$\begin{array}{l} \\ \\ c(x)=\frac{C(x)}{x} \\ c(x)=\frac{10x+3000}{x}\end{array}$
	The average cost function is $c(x)=\frac{10x+3000}{x}.$

ⓑ

We want the function $c(x)$ to be less than $40.$	$\ c(x)<40$
Substitute the rational expression for $c(x).$	$\ \frac{10x+3000}{x}<40\ x\ne 0$
Subtract 40 to get 0 on the right.	$\ \frac{10x+3000}{x}-40<0$
Rewrite the left side as one quotient by finding the LCD and performing the subtraction.	$\frac{10x+3000}{x}-40(\frac{x}{x})<0$
	$\ \frac{10x+3000}{x}-\frac{40x}{x}<0$
	$\ \frac{10x+3000-40x}{x}<0$
	$\ \frac{-30x+3000}{x}<0$
Factor the numerator to show all factors.	$\ \frac{-30(x-100)}{x}<0$
Find the zero partition numbers.	$\begin{array}{llll}-30(x-100) & = & 0 & \ x=0 \\ -30\ne 0\ x-100 & = & 0 & \\ x & = & 100 & \end{array}$

More than 100 items must be produced to keep the average cost below $40 per item.

The function $C(x)=20x+6000$ represents the cost to produce $x,$ number of items. Find ⓐ the average cost function, $c(x)$ ⓑ how many items should be produced so that the average cost is less than $60?

Otkrij odgovor

ⓐ $c(x)=\frac{20x+6000}{x}$
ⓑ More than 150 items must be produced to keep the average cost below $60 per item.
The function $C(x)=5x+900$ represents the cost to produce $x,$ number of items. Find ⓐ the average cost function, $c(x)$ ⓑ how many items should be produced so that the average cost is less than $20?

Otkrij odgovor

ⓐ $c(x)=\frac{5x+900}{x}$ ⓑ More than 60 items must be produced to keep the average cost below $20 per item.
$\frac{x-3}{x+4}\ge 0$

Otkrij odgovor

$(\text{-}\infty ,-4)\cup [3,\infty )$
$\frac{x+6}{x-5}\ge 0$
$\frac{x+1}{x-3}\le 0$

Otkrij odgovor

$[-1,3)$
$\frac{x-4}{x+2}\le 0$
$\frac{x-7}{x-1}>0$

Otkrij odgovor

$(\text{-}\infty ,1)\cup (7,\infty )$
$\frac{x+8}{x+3}>0$
$\frac{x-6}{x+5}<0$

Otkrij odgovor

$(-5,6)$
$\frac{x+5}{x-2}<0$
$\frac{3x}{x-5}<1$

Otkrij odgovor

$(-\frac{5}{2},5)$
$\frac{5x}{x-2}<1$
$\frac{6x}{x-6}>2$

Otkrij odgovor

$(\text{-}\infty ,-3)\cup (6,\infty )$
$\frac{3x}{x-4}>2$
$\frac{2x+3}{x-6}\le 1$

Otkrij odgovor

$[-9,6)$
$\frac{4x-1}{x-4}\le 1$
$\frac{3x-2}{x-4}\ge 2$

Otkrij odgovor

$(\text{-}\infty ,-6]\cup (4,\infty )$
$\frac{4x-3}{x-3}\ge 2$
$\frac{1}{{x}^{2}+7x+12}>0$

Otkrij odgovor

$(\text{-}\infty ,-4)\cup (-3,\infty )$
$\frac{1}{{x}^{2}-4x-12}>0$
$\frac{3}{{x}^{2}-5x+4}<0$

Otkrij odgovor

$(1,4)$

Symbols used here

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Rational Inequalities

Solve rational inequalities
Solve an inequality with rational functions
Write the inequality as one quotient on the left and zero on the right.
Determine the zero partition numbers–the points where the rational expression will be zero or undefined.
Use the zero partition numbers to divide the number line into intervals.
Test a value in each interval. Above the number line show the sign of each factor of the numerator and denominator in each interval. Below the number line show the sign of the quotient.
Determine the intervals where the inequality is correct. Write the solution in interval notation.
Write the inequality as one quotient on the left and zero on the right.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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	\(R(x)\le 0\)
Substitute the rational expression for \(R(x).\)	\(\frac{x+3}{x-5}\le 0\ x\ne 5\)
Find the zero partition numbers.	\(\begin{array}{llllllll}x+3 & = & 0 & & & x-5 & = & 0 \\ x & = & -3 & & & x & = & 5\end{array}\)
Use the zero partition numbers to divide the number line into intervals.

Test values in each interval. Above the number line, show the sign of each factor in each interval. Below the number line, show the sign of the quotient
Write the solution in interval notation. Since 5 is excluded we, do not include it in the interval.	\([-3,5)\)

	\(C(x)=10x+3000\)
The average cost function is \(c(x)=\frac{C(x)}{x}.\)
To find the average cost function, divide the cost function by \(x.\)	\(\begin{array}{l} \\ \\ c(x)=\frac{C(x)}{x} \\ c(x)=\frac{10x+3000}{x}\end{array}\)
	The average cost function is \(c(x)=\frac{10x+3000}{x}.\)

We want the function \(c(x)\) to be less than \(40.\)	\(\ c(x)<40\)
Substitute the rational expression for \(c(x).\)	\(\ \frac{10x+3000}{x}<40\ x\ne 0\)
Subtract 40 to get 0 on the right.	\(\ \frac{10x+3000}{x}-40<0\)
Rewrite the left side as one quotient by finding the LCD and performing the subtraction.	\(\frac{10x+3000}{x}-40(\frac{x}{x})<0\)
	\(\ \frac{10x+3000}{x}-\frac{40x}{x}<0\)
	\(\ \frac{10x+3000-40x}{x}<0\)
	\(\ \frac{-30x+3000}{x}<0\)
Factor the numerator to show all factors.	\(\ \frac{-30(x-100)}{x}<0\)
Find the zero partition numbers.	\(\begin{array}{llll}-30(x-100) & = & 0 & \ x=0 \\ -30\ne 0\ x-100 & = & 0 & \\ x & = & 100 & \end{array}\)

	\(\frac{1}{3}-\frac{2}{{x}^{2}}<\frac{5}{3x}\)
Subtract \(\frac{5}{3x}\) to get zero on the right.	\(\frac{1}{3}-\frac{2}{{x}^{2}}-\frac{5}{3x}<0\)
Rewrite to get each fraction with the LCD \(3{x}^{2}.\)	\(\frac{1⋅{x}^{2}}{3⋅{x}^{2}}-\frac{2⋅3}{{x}^{2}⋅3}-\frac{5⋅x}{3x⋅x}<0\)
Simplify.	\(\frac{{x}^{2}}{3{x}^{2}}-\frac{6}{3{x}^{2}}-\frac{5x}{3{x}^{2}}<0\)
Subtract the numerators and place the difference over the common denominator.	\(\frac{{x}^{2}-5x-6}{3{x}^{2}}<0\)
Factor the numerator.	\(\frac{(x-6)(x+1)}{3{x}^{2}}<0\)
Find the zero partition numbers.	\(\begin{array}{lllllllllllll} \\ 3{x}^{2} & = & 0 & & & x-6 & = & 0 & & & x+1 & = & 0 \\ x & = & 0 & & & x & = & 6 & & & x & = & \text{-}1\end{array}\)
Use the zero partition numbers to divide the number line into intervals.

Above the number line show the sign of each factor in each interval. Below the number line, show the sign of the quotient.
Since, 0 is excluded, the solution is the two intervals, \((-1,0)\) and \((0,6).\)	\((-1,0)\cup (0,6)\)

	\(R(x)\le 0\)
Substitute the rational expression for \(R(x).\)	\(\frac{x+3}{x-5}\le 0\ x\ne 5\)
Find the zero partition numbers.	\(\begin{array}{llllllll}x+3 & = & 0 & & & x-5 & = & 0 \\ x & = & -3 & & & x & = & 5\end{array}\)
Use the zero partition numbers to divide the number line into intervals.

Test values in each interval. Above the number line, show the sign of each factor in each interval. Below the number line, show the sign of the quotient
Write the solution in interval notation. Since 5 is excluded we, do not include it in the interval.	\([-3,5)\)

	\(\frac{4x}{x-6}<1\)
Subtract 1 to get zero on the right.	\(\frac{4x}{x-6}-1<0\)
Rewrite 1 as a fraction using the LCD.	\(\frac{4x}{x-6}-\frac{x-6}{x-6}<0\)
Subtract the numerators and place the difference over the common denominator.	\(\frac{4x-(x-6)}{x-6}<0\)
Simplify.	\(\frac{3x+6}{x-6}<0\)
Factor the numerator to show all factors.	\(\frac{3(x+2)}{x-6}<0\)
Find the zero partition numbers.
The quotient will be zero when the numerator is zero. The quotient is undefined when the denominator is zero.	\(\begin{array}{llllllll}x+2 & = & 0 & & & x-6 & = & 0 \\ x & = & \text{-}2 & & & x & = & 6\end{array}\)
Use the zero partition numbers to divide the number line into intervals.

Test a value in each interval.

Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

Determine the intervals where the inequality is correct. We want the quotient to be negative, so the solution includes the points between −2 and 6. Since the inequality is strictly less than, the endpoints are not included.
We write the solution in interval notation as (−2, 6).

The inequality is in the correct form.	\(\frac{5}{{x}^{2}-2x-15}>0\)
Factor the denominator.	\(\frac{5}{(x+3)(x-5)}>0\)
Find the zero partition numbers. The quotient is 0 when the numerator is 0. Since the numerator is always 5, the quotient cannot be 0.
The quotient will be undefined when the denominator is zero.	\(\begin{array}{l}(x+3)(x-5)=0 \\ x=-3,\ x=5\end{array}\)
Use the zero partition numbers to divide the number line into intervals.

Test values in each interval. Above the number line show the sign of each factor of the denominator in each interval. Below the number line, show the sign of the quotient.
Write the solution in interval notation.	\((\text{-}\infty ,-3)\cup (5,\infty )\)