2.4: Rational Inequalities

Solving rational inequalities involves finding the zeroes of the numerator and denominator, then using these values to investigate solution set regions on the number line.

Example 2.4.1

Solve the inequalities and write the solution sets in interval notation:

1. \(\dfrac{x − 1}{x + 1} ≥ 0\)
2. \(\dfrac{2x − 3}{x + 1} ≤ 0\)
3. \(\dfrac{x + 2}{x − 2} ≥ 0\)

Solution

1. \(\dfrac{x - 1}{x + 1} ≥ 0\) Set x - 1 = 0 and solve  Add 1 to both sides x - 1 + 1 = 0 + 1 x + 0 = 1 x = 1 Set x + 1 = 0 and solve Subtract 1 from both sides x + 1 - 1 = 0 - 1 x + 0 = -1 x = -1 Since x = -1 will cause the rational expression to be undefined x cannot be equal to -1.

   We have three regions, (-\(\infty\), -1), (-1, 1], and [1, \(\infty\)).  Let's look into each region. Select x = -2. \(\dfrac{-2 - 1}{-2 + 1} = \dfrac{-3}{-1} = 3\) Since it is positive, from negative infinity to -1 is a solution to the inequality (-\(\infty\), -1).  Let's now look in interval (-1, 1]. Select x = 0 \(\dfrac{0 - 1}{0 + 1} = \dfrac{-1}{1} < 0\) Since it is negative, (-1, 1] is not a solution to the inequality. \(\dfrac{2 - 1}{2 + 1} = \dfrac{1}{3} ≥ 0\) Since it is positive, [1, oo). (-\(\infty\), -1) U [1, \(\infty\)) is the

   
   
    

The zeroes divide the number line into 3 regions, x < −1, −1 < x < 1, x > 1

 

2. \(\dfrac{2x − 3}{x + 1} ≤ 0\)

   Set the numerator and denominator equal to 0 and solve for x. 2x - 3 = 0 Add 3 to both sides 2x - 3 + 3 = 0 + 3 2x + 0 = 3 2x = 3 Divide by 2 on both sides \( x = \dfrac{3}{2}\) x + 1 = 0 Subtract 1 from both sides x + 1 - 1 = 0 - 1 x + 0 = -1 x = -1 There are three regions, x < -1, -1< x < 3/2, x > 3/2.

   In region x < -1, select x = -2.  Substitute in equation. \(\dfrac{2(-1)-3}{-2 + 1} = \dfrac{-5}{-1} = 5\)   Since the value is positive, it is not a solution for the inequality. In region -1 < x < 3/2, select x = 0.  Substitute in equation. \(\dfrac{2(0) - 3}{0 + 1} = \dfrac{-3}{1} = -3\) Since the value is negative, it is a solution for the inequality. In region x > 3/2, select x = 2.  Substitute in equation. \(\dfrac{2(2) - 3}{2 + 3} = \dfrac{1}{5}\) Since the value is positive, it is not a solution for the inequality. (-1, 3/2] is the solution for the inequality.

    

The zeroes divide the number line into 3 regions, x < −1, −1 < x < 1.5, x > 1.5

 

 

3. \(\dfrac{x + 2}{x − 2} ≥ 0 \)

   Set x + 2 and x - 2 equal to 0 and solve. x + 2 = 0 Subtract 2 on both sides x + 2 - 2 = 0 - 2 x + 0 = -2 x = -2 x - 2 = 0 Add 2 to both sides x - 2 + 2 = 0 + 2 x + 0 = 2 x = 2 There are three regions x < -2, -2 < x < 2, x > 2

   For the region x < -2, select a value x = -3. \(\dfrac{-3 + 2}{-3 - 2} = \dfrac{-1}{-1} = 1\) Since the value is position, this is a solution to the inequality. For the region -2 < x < 2, select a value x = 0. \(\dfrac{0 + 2}{0 - 2} = \dfrac{2}{-2} = -1\) Since the value is negative, this is not a solution. For the region x > 2, select a value x = 3. \(\dfrac{3 + 2}{3 - 2} = \dfrac{5}{1} = 5\) Since the value is positive, this is a solution to the inequality. The solution is (-\(\infty\), -2] U [2, \(\infty\)).

    

The zeroes divide the number line into 3 regions
 x < −2, −2 < x < 2, x > 2

 

 

Exercise 2.4.1

1. \(\dfrac{x + 3}{x − 2} ≥ 0\)
2. \(\dfrac{x − 2}{x − 1} ≤ 0\)
3. \(\dfrac{8}{x + 2}\) ≤\(\dfrac{1}{x + 2}\)

 

Exercise Solutions \(\PageIndex{1}\)

\(\dfrac{x + 3}{x − 2} ≥ 0\)

Solution

Set x + 3 = 0 and solve for x.

x + 3 = 0
   -3      -3
x = -3

Set x - 2 = 0 and solve for x.
x - 2= 0
   2     2
X + 0 = 2
x = 2

Because we have the line under the inequality, ≥, we will include -3 and 2 in the solution.

The number line is divided into three regions.  Left of -3,  between -3 and 2, to the right of 2.  Pick a point in each region.

x = -4 

\(\dfrac{-4+3}{-4-2}\) = \(\dfrac{-1}{-6}\) = \(\dfrac{1}{6}\) ≥ 0.  The area to the left of -3 is a solution.  The interval notation is 

x = 0

\(\dfrac{0+3}{0-2}\) = \(\dfrac{3}{-2}\) < 0.  The area between -3 and 2 is not a solution.

x = 3

\(\dfrac{3+3}{3-2}\) = \(\dfrac{6}{1}\) = 6 ≥ 0.  The area to the right of 2 is a solution.  The interval notation is 

The solution is  in Interval Notation.

2. Example problem \(\dfrac{2x − 3}{x + 1} ≤ 0 \) The quotient must be less than or equal to 0.  2x − 3 = 0, solve for x, x = 1.5.  Find the zeros of the numerator x + 1 = 0, x = −1.  Find the zeroes of the denominator.

The zeroes divide the number line into \(3\) regions, \(x < −1\), \(−1 < x < 1.5\), \(x > 1.5\)

\begin{flalign*}
&\text{For } x < -1, \text{ choose } x = -2. &\\
&\dfrac{2(-2) - 3}{-2 + 1} = \dfrac{-7}{-1} = 7 \geq 0 &
\end{flalign*}

\begin{flalign*}
&\text{For } -1 < x < 1.5, \text{ choose } x = 0. &\\
&\dfrac{2(0) - 3}{0 + 1} = \dfrac{-3}{1} = -3 \leq 0 &
\end{flalign*}

\begin{flalign*} &\text{For } x > 1.5, \text{ choose } x = 2. &\\ &\dfrac{2(2) - 3}{2 + 1} = \dfrac{1}{3} \geq 0 & \end{flalign*} Replacing \(2\) for \(x\) results in the answer \(\dfrac{1}{3}\), which is greater than or equal to \(0\). Therefore, this region \(x > 1.5\) is included in the solution set.

Final Answer is \begin{flalign*} &(-\infty, -1) \cup (1.5, \infty) & \end{flalign*}

 

3. Example problem \(\dfrac{8}{x + 2}\) ≤\(\dfrac{1}{x + 2}\) 
   \begin{flalign*}
   &\text{Subtract } \dfrac{1}{x+2} \text{ from both sides to get } \dfrac{7}{x + 2} \leq 0 &\\
   &\dfrac{7}{x + 2} \leq 0 \quad \text{The quotient must be greater than or equal to } 0. &\\
   &x + 2 = 0 \implies x = -2 \quad \text{Find the zeroes of the denominator}  &
   \end{flalign*}

The zeroes divide the number line into \(2\) regions, \(x < −2\), \(x > -2\)

\begin{flalign*}
&\text{For } x < -2, \text{ choose } x = -3. &\\
&\dfrac{8}{-3 + 2} = \dfrac{8}{-1} = -8 \leq \dfrac{1}{-3 + 2} = \dfrac{1}{-1} = -1 &
\end{flalign*}

Replacing \(-3\) for \(x\) results in a true inequality (\(-8 \leq -1\)). Therefore, this region \(x < -2\) is included in the solution set.

\[ (-\infty, -2) \]