Solving Rational Inequalities: Step-by-Step Examples with Solutions

Solution

Step 1: Identify zeros

• Numerator: \(-3\) (constant, no zeros)
• Denominator zero: \(-x + 4 = 0 \Rightarrow x = 4\)

Step 2: Test intervals

1. Interval \((-∞, 4)\): Test \(x = 0\) \[ \frac{-3}{-0 + 4} = \frac{-3}{4} < 0 \quad \text{(negative)} \]
2. Interval \((4, ∞)\): Test \(x = 5\) \[ \frac{-3}{-5 + 4} = \frac{-3}{-1} = 3 > 0 \quad \text{(positive)} \]

Step 3: Sign table

\(x\)	\(-∞\)	4	\(∞\)
\(\frac{-3}{-x+4}\)	\(-\)	undef	\(+\)

Solution: \((4, ∞)\)

Graphical verification:

Solution

Zeros: Numerator: \(x = 1\), Denominator: \(x = -2\)

Test intervals:

1. \((-∞, -2)\): Test \(x = -3\) \[ \frac{-3 - 1}{-3 + 2} = \frac{-4}{-1} = 4 > 0 \]
2. \((-2, 1)\): Test \(x = 0\) \[ \frac{0 - 1}{0 + 2} = -\frac{1}{2} < 0 \]
3. \((1, ∞)\): Test \(x = 2\) \[ \frac{2 - 1}{2 + 2} = \frac{1}{4} > 0 \]

Sign table:

\(x\)	\(-∞\)	-2	1	\(∞\)
\(\frac{x-1}{x+2}\)	\(+\)	undef	\(-\)	0	\(+\)

Solution: \((-∞, -2) \cup (1, ∞)\)

Solution

Step 1: Move all terms to one side

\[ \frac{2}{x - 3} - \frac{3}{x + 4} \le 0 \]

Step 2: Combine fractions

\[ \frac{2(x+4) - 3(x-3)}{(x-3)(x+4)} \le 0 \] \[ \frac{-x + 17}{(x-3)(x+4)} \le 0 \]

Step 3: Find critical points

• Numerator zero: \(-x + 17 = 0 \Rightarrow x = 17\)
• Denominator zeros: \(x = 3\) and \(x = -4\)

Step 4: Test intervals

1. \((-∞, -4)\): Test \(x = -5\) → \(\frac{22}{8} > 0\)
2. \((-4, 3)\): Test \(x = 0\) → \(-\frac{17}{12} < 0\)
3. \((3, 17)\): Test \(x = 4\) → \(\frac{13}{8} > 0\)
4. \((17, ∞)\): Test \(x = 18\) → \(-\frac{1}{330} < 0\)

Solution including equality at numerator zero: \((-4, 3) \cup [17, ∞)\)

Solution

Step 1: Combine terms

\[ \frac{2x + 1}{x+2} - \frac{x-4}{x - 3} \ge 0 \] \[ \frac{x^2 - 3x + 5}{(x+2)(x-3)} \ge 0 \]

Step 2: Analyze numerator and denominator

• Numerator: \(x^2 - 3x + 5\) has discriminant \((-3)^2 - 4(1)(5) = -11 < 0\), so always positive
• Denominator zeros: \(x = -2\) and \(x = 3\)

Step 3: Test intervals

1. \((-∞, -2)\): Positive
2. \((-2, 3)\): Negative
3. \((3, ∞)\): Positive

Solution: \((-∞, -2) \cup (3, ∞)\)

Solution

Step 1: Factor and combine

\[ \frac{x}{(x-3)(x+1)} - \frac{1}{x + 3} \le 0 \] \[ \frac{5x + 3}{(x-3)(x+1)(x+3)} \le 0 \]

Step 2: Critical points

• Numerator zero: \(x = -\frac{3}{5}\)
• Denominator zeros: \(x = 3, -1, -3\)

Step 3: Sign analysis

Intervals: \((-∞, -3), (-3, -1), (-1, -\frac{3}{5}), (-\frac{3}{5}, 3), (3, ∞)\)

Solution including equality at numerator zero: \((-3, -1) \cup [-\frac{3}{5}, 3)\)

Solution

Step 1: Combine terms

\[ \frac{3x^2}{|x^2-3x-4|} - \frac{1}{3} \ge 0 \] \[ \frac{9x^2 - |x^2-3x-4|}{3|x^2-3x-4|} \ge 0 \]

Step 2: Consider absolute value cases

Denominator zeros: \(x^2-3x-4=0 \Rightarrow x=-1, 4\)

Step 3: Solve numerator

Case 1: \(|x^2-3x-4| = x^2-3x-4\) when \(x \le -1\) or \(x \ge 4\)
Case 2: \(|x^2-3x-4| = -(x^2-3x-4)\) when \(-1 < x < 4\)

Step 4: Critical points from combined analysis: \(x = -1, -\frac{1}{2}, \frac{4}{5}, 4\)

Solution: \((-∞, -1) \cup (-1, -\frac{1}{2}] \cup [\frac{4}{5}, 4) \cup (4, ∞)\)