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

Step 1: Identify zeros (critical points)

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

Step 2: Test intervals

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

Solution: \((4, ∞)\)

Graphical verification:

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

Test intervals:

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

Solution: \((-∞, -2) \cup (1, ∞)\)

Step 1: Move all terms to one side

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

Step 2: Combine fractions

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

Step 3: Find critical points

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

Step 4: Test intervals

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

Note: Because the inequality is "\(\le\)", we include the zeros of the numerator in the solution but NOT the zeros of the denominator (as they make the expression undefined).

Solution: \((-4, 3) \cup [17, ∞)\)

Step 1: Move to one side and combine

\[ \dfrac{2x + 1}{x+2} - \dfrac{x-4}{x - 3} \ge 0 \] \[ \dfrac{(2x+1)(x-3) - (x-4)(x+2)}{(x+2)(x-3)} \ge 0 \] \[ \dfrac{(2x^2 - 5x - 3) - (x^2 - 2x - 8)}{(x+2)(x-3)} \ge 0 \] \[ \dfrac{x^2 - 3x + 5}{(x+2)(x-3)} \ge 0 \]

Step 2: Analyze numerator and denominator

• Numerator: \(x^2 - 3x + 5\). Its discriminant is \(b^2 - 4ac = (-3)^2 - 4(1)(5) = -11\). Since the discriminant is negative and the leading coefficient is positive, the numerator is always positive for all real \(x\).
• Denominator zeros: \(x = -2\) and \(x = 3\)

Step 3: Test intervals

Since the numerator is always positive, the fraction's sign depends entirely on the denominator \((x+2)(x-3)\).

1. \((-∞, -2)\): Denominator is positive. Fraction is \(+\)
2. \((-2, 3)\): Denominator is negative. Fraction is \(-\)
3. \((3, ∞)\): Denominator is positive. Fraction is \(+\)

Solution: \((-∞, -2) \cup (3, ∞)\)

Step 1: Factor and combine

Factor \(x^2 - 2x - 3\) to \((x-3)(x+1)\):

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

Find the common denominator \((x-3)(x+1)(x+3)\):

\[ \dfrac{x(x+3) - 1(x-3)(x+1)}{(x-3)(x+1)(x+3)} \le 0 \] \[ \dfrac{x^2 + 3x - (x^2 - 2x - 3)}{(x-3)(x+1)(x+3)} \le 0 \] \[ \dfrac{5x + 3}{(x-3)(x+1)(x+3)} \le 0 \]

Step 2: Critical points

• Numerator zero: \(5x + 3 = 0 \implies x = -\dfrac{3}{5}\)
• Denominator zeros: \(x = 3, -1, -3\)

Step 3: Sign analysis

Intervals to test: \((-∞, -3), (-3, -1), (-1, -3/5), (-3/5, 3), (3, ∞)\)

By evaluating test points in each interval, the expression is \(\le 0\) in:

Solution: \((-3, -1) \cup [-\dfrac{3}{5}, 3)\)

Step 1: Establish the domain

The denominator cannot be zero. \(x^2 - 3x - 4 = (x-4)(x+1) = 0 \implies x = -1, 4\). The domain is \(x \neq -1, 4\).

Since an absolute value is always positive (for \(x\) in the domain), we can safely multiply both sides by \(3|x^2-3x-4|\) without reversing the inequality:

\[ 9x^2 \ge |x^2-3x-4| \]

Step 2: Consider absolute value cases

Case 1: \(x^2-3x-4 > 0\) (which occurs when \(x < -1\) or \(x > 4\))

\[ 9x^2 \ge x^2 - 3x - 4 \] \[ 8x^2 + 3x + 4 \ge 0 \]

The discriminant of \(8x^2 + 3x + 4\) is \(3^2 - 4(8)(4) = 9 - 128 = -119 < 0\). It is always positive. Thus, all \(x\) in this case's interval satisfy the inequality.
Valid solutions for Case 1: \((-∞, -1) \cup (4, ∞)\)

Case 2: \(x^2-3x-4 < 0\) (which occurs when \(-1 < x < 4\))

\[ 9x^2 \ge -(x^2 - 3x - 4) \] \[ 9x^2 \ge -x^2 + 3x + 4 \] \[ 10x^2 - 3x - 4 \ge 0 \]

Factoring gives \((5x - 4)(2x + 1) \ge 0\), which yields critical points \(x = \dfrac{4}{5}\) and \(x = -\dfrac{1}{2}\).
The parabola opens upwards, meaning the solution to the quadratic is \(x \le -\dfrac{1}{2}\) or \(x \ge \dfrac{4}{5}\).
Intersecting this with the Case 2 interval \((-1, 4)\) gives:
Valid solutions for Case 2: \((-1, -\dfrac{1}{2}] \cup [\dfrac{4}{5}, 4)\)

Step 3: Combine all solutions

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