Solve Rational Inequalities Examples With Solutions

Solve the inequality:

$$ \dfrac{x-2}{x+1} \ge 0 $$

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Solution:

Step 1: Identify the critical points (zeros of numerator and denominator)

Critical points occur where the expression is zero or undefined.

• Numerator: $x - 2 = 0 \implies x = 2$
• Denominator: $x + 1 = 0 \implies x = -1$

So the critical points are: $x = -1$ (undefined) and $x = 2$ (zero).

Step 2: Divide the number line into intervals

Use the critical points to divide the number line:

• Interval 1: $(-\infty, -1)$
• Interval 2: $(-1, 2)$
• Interval 3: $(2, \infty)$

Step 3: Test a point in each interval

Choose a value in each interval and test the sign of the expression.

Interval 1: $x = -2$

$$ \dfrac{-2 - 2}{-2 + 1} = \dfrac{-4}{-1} = 4 > 0 \quad (\text{Positive}) $$

Interval 2: $x = 0$

$$ \dfrac{0 - 2}{0 + 1} = \dfrac{-2}{1} = -2 < 0 \quad (\text{Negative}) $$

Interval 3: $x = 3$

$$ \dfrac{3 - 2}{3 + 1} = \dfrac{1}{4} > 0 \quad (\text{Positive}) $$

Step 4: Include or exclude the critical points

$x = -1$ makes the denominator zero and is therefore excluded.

$x = 2$ makes the numerator zero which satisfies $\ge 0$ and is therefore included.

Final Answer: The solution set of the given inequality is:

$$ (-\infty, -1) \cup [2, \infty) $$

Solve the inequality:

$$ \dfrac{x+1}{x+3} \le 2 $$

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Solution:

Step 1: Rewrite the inequality with zero on one side

$$ \dfrac{x + 1}{x + 3} - 2 \le 0 $$

Rewrite with a common denominator:

$$ \dfrac{x + 1 - 2(x + 3)}{x + 3} \le 0 $$

Simplify the numerator:

$$ \dfrac{x + 1 - 2x - 6}{x + 3} = \dfrac{-x - 5}{x + 3} $$

So the inequality becomes:

$$ \dfrac{-x - 5}{x + 3} \le 0 $$

Step 2: Identify the critical points

• Numerator: $-x - 5 = 0 \implies x = -5$
• Denominator: $x + 3 = 0 \implies x = -3$

So the critical points are: $x = -5$ (zero) and $x = -3$ (undefined).

Step 3: Divide the number line into intervals

• Interval 1: $(-\infty, -5)$
• Interval 2: $(-5, -3)$
• Interval 3: $(-3, \infty)$

Step 4: Test a point in each interval

Interval 1: $x = -6$

$$ \dfrac{-(-6) - 5}{-6 + 3} = \dfrac{6 - 5}{-3} = \dfrac{1}{-3} = -\dfrac{1}{3} < 0 $$

Interval 2: $x = -4$

$$ \dfrac{-(-4) - 5}{-4 + 3} = \dfrac{4 - 5}{-1} = \dfrac{-1}{-1} = 1 > 0 $$

Interval 3: $x = 0$

$$ \dfrac{-0 - 5}{0 + 3} = \dfrac{-5}{3} < 0 $$

Step 5: Include or exclude the critical points

$x = -3$ makes the denominator zero and is therefore excluded.

$x = -5$ makes the numerator zero which satisfies $\le 0$ and is therefore included.

Final Answer: The solution set of the given inequality is:

$$ (-\infty, -5] \cup (-3, \infty) $$

Solve the inequality:

$$ \dfrac{4x^2+5x-9}{x^2-x-6} \ge 0 $$

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Solution:

Step 1: Factor numerator and denominator

$$ 4x^2 + 5x - 9 = (4x + 9)(x - 1) $$

$$ x^2 - x - 6 = (x - 3)(x + 2) $$

So the inequality becomes:

$$ \dfrac{(4x + 9)(x - 1)}{(x - 3)(x + 2)} \ge 0 $$

Step 2: Identify the critical points

Numerator:

• $4x + 9 = 0 \implies x = -\dfrac{9}{4}$
• $x - 1 = 0 \implies x = 1$

Denominator:

• $x - 3 = 0 \implies x = 3$
• $x + 2 = 0 \implies x = -2$

So the critical points are: $x = -\dfrac{9}{4}, x = -2, x = 1, x = 3$.

Step 3: Arrange the critical points on the number line

From left to right on the number line: $-\infty, -\dfrac{9}{4}, -2, 1, 3, +\infty$. So the intervals are:

• Interval 1: $(-\infty, -\dfrac{9}{4})$
• Interval 2: $(-\dfrac{9}{4}, -2)$
• Interval 3: $(-2, 1)$
• Interval 4: $(1, 3)$
• Interval 5: $(3, \infty)$

Step 4: Test a point in each interval

Interval 1: $x = -3$

$$ \dfrac{(4(-3) + 9)(-3 - 1)}{(-3 - 3)(-3 + 2)} = \dfrac{(-3)(-4)}{(-6)(-1)} = \dfrac{12}{6} = 2 > 0 $$

Interval 2: $x = -2.1$

$$ \dfrac{(4(-2.1) + 9)(-2.1 - 1)}{(-2.1 - 3)(-2.1 + 2)} \approx -3.64705\dots < 0 $$

Interval 3: $x = 0$

$$ \dfrac{(4(0) + 9)(0 - 1)}{(0 - 3)(0 + 2)} = \dfrac{(9)(-1)}{(-3)(2)} = \dfrac{-9}{-6} = 1.5 > 0 $$

Interval 4: $x = 2$

$$ \dfrac{(4(2) + 9)(2 - 1)}{(2 - 3)(2 + 2)} = \dfrac{(17)(1)}{(-1)(4)} = \dfrac{17}{-4} = -4.25 < 0 $$

Interval 5: $x = 4$

$$ \dfrac{(4(4) + 9)(4 - 1)}{(4 - 3)(4 + 2)} = \dfrac{(25)(3)}{(1)(6)} = \dfrac{75}{6} > 0 $$

Step 5: Include or exclude the critical points

• $x = -\dfrac{9}{4}$ and $x = 1$ $\to$ zeros of the numerator $\to$ satisfy $\ge 0$ $\to$ included
• $x = -2$ and $x = 3$ $\to$ make the denominator zero $\to$ excluded

Step 6: Determine where the expression is $\ge 0$

• $(-\infty, -\dfrac{9}{4}]$: positive $\to$ included
• $(-\dfrac{9}{4}, -2)$: negative $\to$ excluded
• $(-2, 1]$: positive $\to$ included
• $(1, 3)$: negative $\to$ excluded
• $(3, \infty)$: positive $\to$ included

Final Answer:

$$ (-\infty, -\dfrac{9}{4}] \cup (-2, 1] \cup (3, \infty) $$

Solve the inequality:

$$ \dfrac{x^3+1}{x^2-9} < 0 $$

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Solution:

Step 1: Factor the expression

• Numerator: $x^3 + 1 = (x + 1)(x^2 - x + 1)$
• Denominator: $x^2 - 9 = (x - 3)(x + 3)$

So the inequality becomes:

$$ \dfrac{(x + 1)(x^2 - x + 1)}{(x - 3)(x + 3)} < 0 $$

Step 2: Identify the critical points

• $x + 1 = 0 \implies x = -1$
• $x - 3 = 0 \implies x = 3$
• $x + 3 = 0 \implies x = -3$

Note: $x^2 - x + 1$ has no real roots because its discriminant is negative ($\Delta = (-1)^2 - 4(1)(1) = -3$). Furthermore, because the leading coefficient is positive, this term is always positive for all real $x$ and will not affect the sign chart.

So the critical points are: $x = -3$ (undefined), $x = -1$ (zero), $x = 3$ (undefined).

Step 3: Divide the number line into intervals

• Interval 1: $(-\infty, -3)$
• Interval 2: $(-3, -1)$
• Interval 3: $(-1, 3)$
• Interval 4: $(3, \infty)$

Step 4: Test a point in each interval

Interval 1: $x = -4$

$$ \dfrac{(-3)(21)}{(-7)(-1)} = \dfrac{-63}{7} = -9 < 0 $$

Interval 2: $x = -2$

$$ \dfrac{(-1)(7)}{(-5)(1)} = \dfrac{-7}{-5} = \dfrac{7}{5} > 0 $$

Interval 3: $x = 0$

$$ \dfrac{(1)(1)}{(-3)(3)} = \dfrac{1}{-9} < 0 $$

Interval 4: $x = 4$

$$ \dfrac{(5)(13)}{(1)(7)} = \dfrac{65}{7} > 0 $$

Step 5: Include or exclude the critical points

$x = -3$ and $x = 3$ make the denominator zero and are therefore excluded.

$x = -1$ makes the numerator zero which does not satisfy $< 0$ (strict inequality) and is therefore excluded.

Final Answer: The solution set of the given inequality is:

$$ (-\infty, -3) \cup (-1, 3) $$

Solve graphically:

$$ \dfrac{x + 2}{x - 1} \ge \ln(x + 1) $$

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Solution:

Graph the two sides of the inequality and approximate the coordinates of the points of intersection of the two graphs; then determine the intervals over which $\dfrac{x + 2}{x - 1}$ is greater than or equal to $\ln(x + 1)$.

$\ln(x + 1)$ is defined for $x > -1$, hence any solution set must satisfy $x > -1$. The function $\dfrac{x + 2}{x - 1}$ has a vertical asymptote at $x = 1$.

From the graphs, $\dfrac{x + 2}{x - 1}$ (green) is greater than or equal to $\ln(x + 1)$ (red) for:

$$ (-1, -0.59] \cup (1, 4.88] $$

Solve the double rational inequality:

$$ -1 < \dfrac{2x - 1}{x + 2} \le 3 $$

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Solution:

A double inequality must be solved by splitting it into two separate inequalities. The final solution is the intersection (overlap) of the two solution sets.

Part 1: Left Inequality

$$ -1 < \dfrac{2x - 1}{x + 2} \implies 0 < \dfrac{2x - 1}{x + 2} + 1 $$

Find a common denominator:

$$ 0 < \dfrac{2x - 1 + x + 2}{x + 2} \implies \dfrac{3x + 1}{x + 2} > 0 $$

Critical points: $x = -\dfrac{1}{3}$ and $x = -2$.

Sign chart for $(-\infty, -2)$, $(-2, -1/3)$, $(-1/3, \infty)$ yields positive results on the outer intervals.

Solution set 1: $(-\infty, -2) \cup (-\dfrac{1}{3}, \infty)$

Part 2: Right Inequality

$$ \dfrac{2x - 1}{x + 2} \le 3 \implies \dfrac{2x - 1}{x + 2} - 3 \le 0 $$

Find a common denominator:

$$ \dfrac{2x - 1 - 3(x + 2)}{x + 2} \le 0 \implies \dfrac{-x - 7}{x + 2} \le 0 $$

Critical points: $x = -7$ and $x = -2$.

Sign chart for $(-\infty, -7)$, $(-7, -2)$, $(-2, \infty)$ yields negative results on the outer intervals. We include $-7$ because of the $\le$ sign.

Solution set 2: $(-\infty, -7] \cup (-2, \infty)$

Intersection:

We must find where Solution 1 AND Solution 2 overlap.

• The left sides overlap from $(-\infty, -7]$.
• The right sides overlap from $(-\dfrac{1}{3}, \infty)$.

Final Answer:

$$ (-\infty, -7] \cup \left(-\dfrac{1}{3}, \infty\right) $$

Solve the inequality:

$$ \dfrac{(x - 3)^2 (x + 1)}{(x - 2)^3} \le 0 $$

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Solution:

This problem introduces the concept of multiplicity. When a factor is squared (or has any even power), the sign of the expression will not change when crossing that critical point. When a factor has an odd power, the sign will change.

Step 1: Identify Critical Points

• Numerator: $x = 3$ (even multiplicity), $x = -1$ (odd multiplicity)
• Denominator: $x = 2$ (odd multiplicity, excluded)

Step 2: Sign Chart

Intervals to test: $(-\infty, -1), (-1, 2), (2, 3), (3, \infty)$

• Test $x = -2$: $\dfrac{(+)(-)}{(-)} = (+) > 0$
• Test $x = 0$: $\dfrac{(+)(+)}{(-)} = (-) \le 0$ (This interval works)
• Test $x = 2.5$: $\dfrac{(+)(+)}{(+)} = (+) > 0$
• Test $x = 4$: $\dfrac{(+)(+)}{(+)} = (+) > 0$ (Notice the sign didn't change crossing $x=3$)

Step 3: Analyze Endpoints

We need where the function is negative OR exactly zero.

• The interval $(-1, 2)$ is negative.
• $x = -1$ makes it zero (Included).
• $x = 2$ makes it undefined (Excluded).
• $x = 3$ makes it zero (Included).

Even though the interval around $3$ is positive, the exact point $x=3$ yields $0$, which satisfies $\le 0$. This creates an isolated solution point.

Final Answer:

$$ [-1, 2) \cup \{3\} $$