PPT - Solving Equations Containing Rational Expressions PowerPoint ... - Free Printable

Problem: Solving Rational Equations & Inequalities Practice

We are tasked with solving the following rational equations and inequalities. Let's solve each one step by step.

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#### 1. Solve the equation:
\[
\frac{2x}{6x+5} = \frac{x+3}{3x-1}
\]

Step 1: Eliminate the denominators by cross-multiplying.
\[
2x(3x - 1) = (x + 3)(6x + 5)
\]

Step 2: Expand both sides.
\[
2x(3x - 1) = 6x^2 - 2x
\]
\[
(x + 3)(6x + 5) = 6x^2 + 5x + 18x + 15 = 6x^2 + 23x + 15
\]

So the equation becomes:
\[
6x^2 - 2x = 6x^2 + 23x + 15
\]

Step 3: Simplify by subtracting \(6x^2\) from both sides.
\[
-2x = 23x + 15
\]

Step 4: Combine like terms.
\[
-2x - 23x = 15
\]
\[
-25x = 15
\]

Step 5: Solve for \(x\).
\[
x = -\frac{15}{25} = -\frac{3}{5}
\]

Step 6: Check for extraneous solutions.
The original denominators are \(6x + 5\) and \(3x - 1\). Substitute \(x = -\frac{3}{5}\) into these denominators:
\[
6\left(-\frac{3}{5}\right) + 5 = -\frac{18}{5} + 5 = -\frac{18}{5} + \frac{25}{5} = \frac{7}{5} \neq 0
\]
\[
3\left(-\frac{3}{5}\right) - 1 = -\frac{9}{5} - 1 = -\frac{9}{5} - \frac{5}{5} = -\frac{14}{5} \neq 0
\]
Since neither denominator is zero, \(x = -\frac{3}{5}\) is a valid solution.

Final Answer:
\[
\boxed{-\frac{3}{5}}
\]

---

#### 2. Solve the inequality:
\[
\frac{x^2 + 9x + 14}{x^2 - 6x + 5} > 0
\]

Step 1: Factor the numerator and denominator.
\[
x^2 + 9x + 14 = (x + 2)(x + 7)
\]
\[
x^2 - 6x + 5 = (x - 1)(x - 5)
\]

So the inequality becomes:
\[
\frac{(x + 2)(x + 7)}{(x - 1)(x - 5)} > 0
\]

Step 2: Determine the critical points.
The critical points are the values of \(x\) that make the numerator or denominator zero:
\[
x + 2 = 0 \implies x = -2
\]
\[
x + 7 = 0 \implies x = -7
\]
\[
x - 1 = 0 \implies x = 1
\]
\[
x - 5 = 0 \implies x = 5
\]

These points divide the real number line into intervals:
\[
(-\infty, -7), (-7, -2), (-2, 1), (1, 5), (5, \infty)
\]

Step 3: Test the sign of the expression in each interval.
Choose a test point in each interval and evaluate the sign of \(\frac{(x + 2)(x + 7)}{(x - 1)(x - 5)}\).

- For \(x \in (-\infty, -7)\), choose \(x = -8\):
\[
\frac{(-8 + 2)(-8 + 7)}{(-8 - 1)(-8 - 5)} = \frac{(-6)(-1)}{(-9)(-13)} = \frac{6}{117} > 0
\]

- For \(x \in (-7, -2)\), choose \(x = -5\):
\[
\frac{(-5 + 2)(-5 + 7)}{(-5 - 1)(-5 - 5)} = \frac{(-3)(2)}{(-6)(-10)} = \frac{-6}{60} < 0
\]

- For \(x \in (-2, 1)\), choose \(x = 0\):
\[
\frac{(0 + 2)(0 + 7)}{(0 - 1)(0 - 5)} = \frac{(2)(7)}{(-1)(-5)} = \frac{14}{5} > 0
\]

- For \(x \in (1, 5)\), choose \(x = 3\):
\[
\frac{(3 + 2)(3 + 7)}{(3 - 1)(3 - 5)} = \frac{(5)(10)}{(2)(-2)} = \frac{50}{-4} < 0
\]

- For \(x \in (5, \infty)\), choose \(x = 6\):
\[
\frac{(6 + 2)(6 + 7)}{(6 - 1)(6 - 5)} = \frac{(8)(13)}{(5)(1)} = \frac{104}{5} > 0
\]

Step 4: Determine the solution set.
The expression is positive in the intervals \((- \infty, -7)\), \((-2, 1)\), and \((5, \infty)\). Since the inequality is strict (\(>\)), we do not include the critical points.

Final Answer:
\[
\boxed{(-\infty, -7) \cup (-2, 1) \cup (5, \infty)}
\]

---

#### 3. Solve the inequality:
\[
\frac{x + 3}{x - 4} \geq 0
\]

Step 1: Determine the critical points.
The critical points are the values of \(x\) that make the numerator or denominator zero:
\[
x + 3 = 0 \implies x = -3
\]
\[
x - 4 = 0 \implies x = 4
\]

These points divide the real number line into intervals:
\[
(-\infty, -3), (-3, 4), (4, \infty)
\]

Step 2: Test the sign of the expression in each interval.
Choose a test point in each interval and evaluate the sign of \(\frac{x + 3}{x - 4}\).

- For \(x \in (-\infty, -3)\), choose \(x = -4\):
\[
\frac{-4 + 3}{-4 - 4} = \frac{-1}{-8} > 0
\]

- For \(x \in (-3, 4)\), choose \(x = 0\):
\[
\frac{0 + 3}{0 - 4} = \frac{3}{-4} < 0
\]

- For \(x \in (4, \infty)\), choose \(x = 5\):
\[
\frac{5 + 3}{5 - 4} = \frac{8}{1} > 0
\]

Step 3: Determine the solution set.
The expression is non-negative in the intervals \((- \infty, -3]\) and \([4, \infty)\). Since the inequality is non-strict (\(\geq\)), we include the critical points where the expression is zero or undefined (but not where it is undefined).

Final Answer:
\[
\boxed{(-\infty, -3] \cup (4, \infty)}
\]

---

#### 4. Solve the inequality:
\[
\frac{5}{x - 3} \geq \frac{3}{x - 2}
\]

Step 1: Combine the fractions into a single inequality.
\[
\frac{5}{x - 3} - \frac{3}{x - 2} \geq 0
\]

Step 2: Find a common denominator.
The common denominator is \((x - 3)(x - 2)\):
\[
\frac{5(x - 2) - 3(x - 3)}{(x - 3)(x - 2)} \geq 0
\]

Step 3: Simplify the numerator.
\[
5(x - 2) - 3(x - 3) = 5x - 10 - 3x + 9 = 2x - 1
\]

So the inequality becomes:
\[
\frac{2x - 1}{(x - 3)(x - 2)} \geq 0
\]

Step 4: Determine the critical points.
The critical points are the values of \(x\) that make the numerator or denominator zero:
\[
2x - 1 = 0 \implies x = \frac{1}{2}
\]
\[
x - 3 = 0 \implies x = 3
\]
\[
x - 2 = 0 \implies x = 2
\]

These points divide the real number line into intervals:
\[
(-\infty, \frac{1}{2}), \left(\frac{1}{2}, 2\right), (2, 3), (3, \infty)
\]

Step 5: Test the sign of the expression in each interval.
Choose a test point in each interval and evaluate the sign of \(\frac{2x - 1}{(x - 3)(x - 2)}\).

- For \(x \in (-\infty, \frac{1}{2})\), choose \(x = 0\):
\[
\frac{2(0) - 1}{(0 - 3)(0 - 2)} = \frac{-1}{(-3)(-2)} = \frac{-1}{6} < 0
\]

- For \(x \in \left(\frac{1}{2}, 2\right)\), choose \(x = 1\):
\[
\frac{2(1) - 1}{(1 - 3)(1 - 2)} = \frac{1}{(-2)(-1)} = \frac{1}{2} > 0
\]

- For \(x \in (2, 3)\), choose \(x = 2.5\):
\[
\frac{2(2.5) - 1}{(2.5 - 3)(2.5 - 2)} = \frac{5 - 1}{(-0.5)(0.5)} = \frac{4}{-0.25} < 0
\]

- For \(x \in (3, \infty)\), choose \(x = 4\):
\[
\frac{2(4) - 1}{(4 - 3)(4 - 2)} = \frac{8 - 1}{(1)(2)} = \frac{7}{2} > 0
\]

Step 6: Determine the solution set.
The expression is non-negative in the intervals \(\left[\frac{1}{2}, 2\right)\) and \((3, \infty)\). Since the inequality is non-strict (\(\geq\)), we include the critical point where the expression is zero (\(x = \frac{1}{2}\)) but exclude the points where the expression is undefined (\(x = 2\) and \(x = 3\)).

Final Answer:
\[
\boxed{\left[\frac{1}{2}, 2\right) \cup (3, \infty)}
\]

---

#### 5. Solve the inequality:
\[
\frac{5}{x + 4} \leq \frac{2}{x + 1}
\]

Step 1: Combine the fractions into a single inequality.
\[
\frac{5}{x + 4} - \frac{2}{x + 1} \leq 0
\]

Step 2: Find a common denominator.
The common denominator is \((x + 4)(x + 1)\):
\[
\frac{5(x + 1) - 2(x + 4)}{(x + 4)(x + 1)} \leq 0
\]

Step 3: Simplify the numerator.
\[
5(x + 1) - 2(x + 4) = 5x + 5 - 2x - 8 = 3x - 3
\]

So the inequality becomes:
\[
\frac{3x - 3}{(x + 4)(x + 1)} \leq 0
\]

Step 4: Factor the numerator.
\[
3x - 3 = 3(x - 1)
\]

So the inequality becomes:
\[
\frac{3(x - 1)}{(x + 4)(x + 1)} \leq 0
\]

Step 5: Determine the critical points.
The critical points are the values of \(x\) that make the numerator or denominator zero:
\[
x - 1 = 0 \implies x = 1
\]
\[
x + 4 = 0 \implies x = -4
\]
\[
x + 1 = 0 \implies x = -1
\]

These points divide the real number line into intervals:
\[
(-\infty, -4), (-4, -1), (-1, 1), (1, \infty)
\]

Step 6: Test the sign of the expression in each interval.
Choose a test point in each interval and evaluate the sign of \(\frac{3(x - 1)}{(x + 4)(x + 1)}\).

- For \(x \in (-\infty, -4)\), choose \(x = -5\):
\[
\frac{3(-5 - 1)}{(-5 + 4)(-5 + 1)} = \frac{3(-6)}{(-1)(-4)} = \frac{-18}{4} < 0
\]

- For \(x \in (-4, -1)\), choose \(x = -2\):
\[
\frac{3(-2 - 1)}{(-2 + 4)(-2 + 1)} = \frac{3(-3)}{(2)(-1)} = \frac{-9}{-2} > 0
\]

- For \(x \in (-1, 1)\), choose \(x = 0\):
\[
\frac{3(0 - 1)}{(0 + 4)(0 + 1)} = \frac{3(-1)}{(4)(1)} = \frac{-3}{4} < 0
\]

- For \(x \in (1, \infty)\), choose \(x = 2\):
\[
\frac{3(2 - 1)}{(2 + 4)(2 + 1)} = \frac{3(1)}{(6)(3)} = \frac{3}{18} > 0
\]

Step 7: Determine the solution set.
The expression is non-positive in the intervals \((- \infty, -4)\), \((-1, 1]\). Since the inequality is non-strict (\(\leq\)), we include the critical point where the expression is zero (\(x = 1\)) but exclude the points where the expression is undefined (\(x = -4\) and \(x = -1\)).

Final Answer:
\[
\boxed{(-\infty, -4) \cup (-1, 1]}
\]

---

#### 6. Solve the inequality:
\[
\frac{x^2 + 5x - 6}{x - 4} \leq 0
\]

Step 1: Factor the numerator.
\[
x^2 + 5x - 6 = (x + 6)(x - 1)
\]

So the inequality becomes:
\[
\frac{(x + 6)(x - 1)}{x - 4} \leq 0
\]

Step 2: Determine the critical points.
The critical points are the values of \(x\) that make the numerator or denominator zero:
\[
x + 6 = 0 \implies x = -6
\]
\[
x - 1 = 0 \implies x = 1
\]
\[
x - 4 = 0 \implies x = 4
\]

These points divide the real number line into intervals:
\[
(-\infty, -6), (-6, 1), (1, 4), (4, \infty)
\]

Step 3: Test the sign of the expression in each interval.
Choose a test point in each interval and evaluate the sign of \(\frac{(x + 6)(x - 1)}{x - 4}\).

- For \(x \in (-\infty, -6)\), choose \(x = -7\):
\[
\frac{(-7 + 6)(-7 - 1)}{-7 - 4} = \frac{(-1)(-8)}{-11} = \frac{8}{-11} < 0
\]

- For \(x \in (-6, 1)\), choose \(x = 0\):
\[
\frac{(0 + 6)(0 - 1)}{0 - 4} = \frac{(6)(-1)}{-4} = \frac{-6}{-4} > 0
\]

- For \(x \in (1, 4)\), choose \(x = 2\):
\[
\frac{(2 + 6)(2 - 1)}{2 - 4} = \frac{(8)(1)}{-2} = \frac{8}{-2} < 0
\]

- For \(x \in (4, \infty)\), choose \(x = 5\):
\[
\frac{(5 + 6)(5 - 1)}{5 - 4} = \frac{(11)(4)}{1} = 44 > 0
\]

Step 4: Determine the solution set.
The expression is non-positive in the intervals \((- \infty, -6]\), \([1, 4)\). Since the inequality is non-strict (\(\leq\)), we include the critical points where the expression is zero (\(x = -6\) and \(x = 1\)) but exclude the point where the expression is undefined (\(x = 4\)).

Final Answer:
\[
\boxed{(-\infty, -6] \cup [1, 4)}
\]

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Final Answers:

1. \(\boxed{-\frac{3}{5}}\)
2. \(\boxed{(-\infty, -7) \cup (-2, 1) \cup (5, \infty)}\)
3. \(\boxed{(-\infty, -3] \cup (4, \infty)}\)
4. \(\boxed{\left[\frac{1}{2}, 2\right) \cup (3, \infty)}\)
5. \(\boxed{(-\infty, -4) \cup (-1, 1]}\)
6. \(\boxed{(-\infty, -6] \cup [1, 4)}\)