Question

(ⅰ) $\frac{x-1}{x+1} \geq 0$ (ii) $\frac{x-2}{x+2} \geq 0$ (iii) $\frac{x-1}{x+1} \geq 1$ (iv) $\frac{x-2}{x+5} > 2$ (v) $2x-3 \leq 5$, $\frac{2x+5}{x+7} \geq 3$ (vi) $\frac{2x+1}{7x-1} > 5$, $\frac{x+7}{x-8} > 2$ (vii) $\frac{5-2x}{3} \leq \frac{x}{6} - 5$ (viii) $\frac{x+7}{x-4} > 2$, $\frac{2x+1}{x-2} > 3$ (ix) $\frac{1}{2}(\frac{3}{5}x+4) > \frac{1}{3}(x-6)$ (x) $\frac{2-3x}{5} < \frac{1-x}{3} < \frac{3+4x}{2}$ (xi) $x + 5 > 2(x + 1)$, $2 - x < 3(x + 2)$. (xii) $2(x - 6) < 3x - 7$, $11 - 2x < 6 - x$. (xiii) $-2 - \frac{x}{4} \leq \frac{1+x}{3}$, $3 - x < 4(x - 3)$

Answer 1

(i) $x \in (-\infty, -1) \cup [1, \infty)$ (ii) $x \in (-\infty, -2) \cup [2, \infty)$ (iii) $x \in (-\infty, -1)$ (iv) $x \in (-12, -5)$ (v) $x \in [-16, -7)$ (vi) No solution ($\emptyset$) (vii) $x \in [8, \infty)$ (viii) $x \in (4, 7)$ (ix) $x \in (-\infty, 120)$ (x) $x \in (1/4, \infty)$ (xi) $x \in (-1, 3)$ (xii) $x \in (5, \infty)$ (xiii) $x \in (3, \infty)$

Answer 2

Here are the solutions for each inequality:

(i) $\frac{x-1}{x+1} \geq 0$

Critical points are $x=1$ (from numerator) and $x=-1$ (from denominator). We test intervals:

• For $x < -1$ (e.g., $x=-2$): $\frac{-2-1}{-2+1} = \frac{-3}{-1} = 3 \geq 0$. (True)
• For $-1 < x < 1$ (e.g., $x=0$): $\frac{0-1}{0+1} = \frac{-1}{1} = -1 < 0$. (False)
• For $x \geq 1$ (e.g., $x=2$): $\frac{2-1}{2+1} = \frac{1}{3} \geq 0$. (True)

Since the denominator cannot be zero, $x \neq -1$. Solution: $x \in (-\infty, -1) \cup [1, \infty)$

(ii) $\frac{x-2}{x+2} \geq 0$

Critical points are $x=2$ and $x=-2$. We test intervals:

• For $x < -2$ (e.g., $x=-3$): $\frac{-3-2}{-3+2} = \frac{-5}{-1} = 5 \geq 0$. (True)
• For $-2 < x < 2$ (e.g., $x=0$): $\frac{0-2}{0+2} = \frac{-2}{2} = -1 < 0$. (False)
• For $x \geq 2$ (e.g., $x=3$): $\frac{3-2}{3+2} = \frac{1}{5} \geq 0$. (True)

Since $x \neq -2$. Solution: $x \in (-\infty, -2) \cup [2, \infty)$

(iii) $\frac{x-1}{x+1} \geq 1$

$\frac{x-1}{x+1} - 1 \geq 0$ $\frac{x-1-(x+1)}{x+1} \geq 0$ $\frac{-2}{x+1} \geq 0$

For this inequality to hold, the denominator $x+1$ must be negative (since the numerator is negative and fixed at -2). $x+1 < 0 \implies x < -1$. Solution: $x \in (-\infty, -1)$

(iv) $\frac{x-2}{x+5} > 2$

$\frac{x-2}{x+5} - 2 > 0$ $\frac{x-2-2(x+5)}{x+5} > 0$ $\frac{x-2-2x-10}{x+5} > 0$ $\frac{-x-12}{x+5} > 0$

Multiply by -1 and reverse the inequality sign: $\frac{x+12}{x+5} < 0$

Critical points are $x=-12$ and $x=-5$. We test intervals:

• For $x < -12$ (e.g., $x=-13$): $\frac{-13+12}{-13+5} = \frac{-1}{-8} = \frac{1}{8} \not< 0$. (False)
• For $-12 < x < -5$ (e.g., $x=-10$): $\frac{-10+12}{-10+5} = \frac{2}{-5} = -\frac{2}{5} < 0$. (True)
• For $x > -5$ (e.g., $x=0$): $\frac{0+12}{0+5} = \frac{12}{5} \not< 0$. (False)

Solution: $x \in (-12, -5)$

(v) $2x-3 \leq 5$ and $\frac{2x+5}{x+7} \geq 3$

Inequality 1: $2x-3 \leq 5$ $2x \leq 8 \implies x \leq 4$. Solution 1: $(-\infty, 4]$

Inequality 2: $\frac{2x+5}{x+7} \geq 3$ $\frac{2x+5}{x+7} - 3 \geq 0$ $\frac{2x+5-3(x+7)}{x+7} \geq 0$ $\frac{2x+5-3x-21}{x+7} \geq 0$ $\frac{-x-16}{x+7} \geq 0$

Multiply by -1 and reverse the inequality sign: $\frac{x+16}{x+7} \leq 0$

Critical points are $x=-16$ and $x=-7$. We test intervals:

• For $x \leq -16$ (e.g., $x=-20$): $\frac{-20+16}{-20+7} = \frac{-4}{-13} = \frac{4}{13} \not\leq 0$. (False)
• For $-16 \leq x < -7$ (e.g., $x=-10$): $\frac{-10+16}{-10+7} = \frac{6}{-3} = -2 \leq 0$. (True)
• For $x > -7$ (e.g., $x=0$): $\frac{0+16}{0+7} = \frac{16}{7} \not\leq 0$. (False)

Solution 2: $[-16, -7)$

Intersection of Solution 1 and Solution 2: $(-\infty, 4] \cap [-16, -7) = [-16, -7)$. Solution: $x \in [-16, -7)$

(vi) $\frac{2x+1}{7x-1} > 5$ and $\frac{x+7}{x-8} > 2$

Inequality 1: $\frac{2x+1}{7x-1} > 5$ $\frac{2x+1}{7x-1} - 5 > 0$ $\frac{2x+1-5(7x-1)}{7x-1} > 0$ $\frac{2x+1-35x+5}{7x-1} > 0$ $\frac{-33x+6}{7x-1} > 0$

Multiply by -1/3 and reverse the inequality sign: $\frac{11x-2}{7x-1} < 0$

Critical points are $x=2/11$ and $x=1/7$. Since $1/7 \approx 0.14$ and $2/11 \approx 0.18$, $1/7 < 2/11$. We test intervals:

• For $x < 1/7$ (e.g., $x=0$): $\frac{-2}{-1} = 2 \not< 0$. (False)
• For $1/7 < x < 2/11$ (e.g., $x=0.15$): $\frac{11(0.15)-2}{7(0.15)-1} = \frac{1.65-2}{1.05-1} = \frac{-0.35}{0.05} = -7 < 0$. (True)
• For $x > 2/11$ (e.g., $x=1$): $\frac{11-2}{7-1} = \frac{9}{6} = 1.5 \not< 0$. (False)

Solution 1: $(1/7, 2/11)$

Inequality 2: $\frac{x+7}{x-8} > 2$ $\frac{x+7}{x-8} - 2 > 0$ $\frac{x+7-2(x-8)}{x-8} > 0$ $\frac{x+7-2x+16}{x-8} > 0$ $\frac{-x+23}{x-8} > 0$

Multiply by -1 and reverse the inequality sign: $\frac{x-23}{x-8} < 0$

Critical points are $x=23$ and $x=8$. We test intervals:

• For $x < 8$ (e.g., $x=0$): $\frac{-23}{-8} = \frac{23}{8} \not< 0$. (False)
• For $8 < x < 23$ (e.g., $x=10$): $\frac{10-23}{10-8} = \frac{-13}{2} = -6.5 < 0$. (True)
• For $x > 23$ (e.g., $x=24$): $\frac{24-23}{24-8} = \frac{1}{16} \not< 0$. (False)

Solution 2: $(8, 23)$

Intersection of Solution 1 and Solution 2: $(1/7, 2/11) \cap (8, 23)$. Since $2/11 \approx 0.18$ and $8$, there is no overlap. Solution: $\emptyset$ (No solution)

(vii) $\frac{5-2x}{3} \leq \frac{x}{6} - 5$

Multiply by the LCM of 3 and 6, which is 6: $6 \cdot \frac{5-2x}{3} \leq 6 \cdot (\frac{x}{6} - 5)$ $2(5-2x) \leq x - 30$ $10 - 4x \leq x - 30$ $10 + 30 \leq x + 4x$ $40 \leq 5x$ $x \geq \frac{40}{5}$ $x \geq 8$ Solution: $x \in [8, \infty)$

(viii) $\frac{x+7}{x-4} > 2$ and $\frac{2x+1}{x-2} > 3$

Inequality 1: $\frac{x+7}{x-4} > 2$ $\frac{x+7}{x-4} - 2 > 0$ $\frac{x+7-2(x-4)}{x-4} > 0$ $\frac{x+7-2x+8}{x-4} > 0$ $\frac{-x+15}{x-4} > 0$

Multiply by -1 and reverse the inequality sign: $\frac{x-15}{x-4} < 0$

Critical points are $x=15$ and $x=4$. We test intervals:

• For $x < 4$ (e.g., $x=0$): $\frac{-15}{-4} = \frac{15}{4} \not< 0$. (False)
• For $4 < x < 15$ (e.g., $x=5$): $\frac{5-15}{5-4} = \frac{-10}{1} = -10 < 0$. (True)
• For $x > 15$ (e.g., $x=20$): $\frac{20-15}{20-4} = \frac{5}{16} \not< 0$. (False)

Solution 1: $(4, 15)$

Inequality 2: $\frac{2x+1}{x-2} > 3$ $\frac{2x+1}{x-2} - 3 > 0$ $\frac{2x+1-3(x-2)}{x-2} > 0$ $\frac{2x+1-3x+6}{x-2} > 0$ $\frac{-x+7}{x-2} > 0$

Multiply by -1 and reverse the inequality sign: $\frac{x-7}{x-2} < 0$

Critical points are $x=7$ and $x=2$. We test intervals:

• For $x < 2$ (e.g., $x=0$): $\frac{-7}{-2} = \frac{7}{2} \not< 0$. (False)
• For $2 < x < 7$ (e.g., $x=3$): $\frac{3-7}{3-2} = \frac{-4}{1} = -4 < 0$. (True)
• For $x > 7$ (e.g., $x=8$): $\frac{8-7}{8-2} = \frac{1}{6} \not< 0$. (False)

Solution 2: $(2, 7)$

Intersection of Solution 1 and Solution 2: $(4, 15) \cap (2, 7) = (4, 7)$. Solution: $x \in (4, 7)$

(ix) $\frac{1}{2}(\frac{3}{5}x+4) > \frac{1}{3}(x-6)$ $\frac{3}{10}x + 2 > \frac{1}{3}x - 2$ $2+2 > \frac{1}{3}x - \frac{3}{10}x$ $4 > (\frac{10-9}{30})x$ $4 > \frac{1}{30}x$

Multiply by 30: $120 > x$ Solution: $x \in (-\infty, 120)$

(x) $\frac{2-3x}{5} < \frac{1-x}{3} < \frac{3+4x}{2}$

This is a system of two inequalities: Inequality 1: $\frac{2-3x}{5} < \frac{1-x}{3}$ Multiply by 15: $3(2-3x) < 5(1-x)$ $6-9x < 5-5x$ $6-5 < -5x+9x$ $1 < 4x$ $x > \frac{1}{4}$ Solution 1: $(1/4, \infty)$

Inequality 2: $\frac{1-x}{3} < \frac{3+4x}{2}$ Multiply by 6: $2(1-x) < 3(3+4x)$ $2-2x < 9+12x$ $2-9 < 12x+2x$ $-7 < 14x$ $x > \frac{-7}{14}$ $x > -\frac{1}{2}$ Solution 2: $(-1/2, \infty)$

Intersection of Solution 1 and Solution 2: $(1/4, \infty) \cap (-1/2, \infty) = (1/4, \infty)$. Solution: $x \in (1/4, \infty)$

(xi) $x + 5 > 2(x + 1)$ and $2 - x < 3(x + 2)$

Inequality 1: $x + 5 > 2(x + 1)$ $x + 5 > 2x + 2$ $5 - 2 > 2x - x$ $3 > x$ Solution 1: $(-\infty, 3)$

Inequality 2: $2 - x < 3(x + 2)$ $2 - x < 3x + 6$ $2 - 6 < 3x + x$ $-4 < 4x$ $x > -1$ Solution 2: $(-1, \infty)$

Intersection of Solution 1 and Solution 2: $(-\infty, 3) \cap (-1, \infty) = (-1, 3)$. Solution: $x \in (-1, 3)$

(xii) $2(x - 6) < 3x - 7$ and $11 - 2x < 6 - x$

Inequality 1: $2(x - 6) < 3x - 7$ $2x - 12 < 3x - 7$ $-12 + 7 < 3x - 2x$ $-5 < x$ Solution 1: $(-5, \infty)$

Inequality 2: $11 - 2x < 6 - x$ $11 - 6 < -x + 2x$ $5 < x$ Solution 2: $(5, \infty)$

Intersection of Solution 1 and Solution 2: $(-5, \infty) \cap (5, \infty) = (5, \infty)$. Solution: $x \in (5, \infty)$

(xiii) $-2 - \frac{x}{4} \leq \frac{1+x}{3}$ and $3 - x < 4(x - 3)$

Inequality 1: $-2 - \frac{x}{4} \leq \frac{1+x}{3}$ Multiply by 12: $12(-2) - 12(\frac{x}{4}) \leq 12(\frac{1+x}{3})$ $-24 - 3x \leq 4(1+x)$ $-24 - 3x \leq 4 + 4x$ $-24 - 4 \leq 4x + 3x$ $-28 \leq 7x$ $x \geq -4$ Solution 1: $[-4, \infty)$

Inequality 2: $3 - x < 4(x - 3)$ $3 - x < 4x - 12$ $3 + 12 < 4x + x$ $15 < 5x$ $x > 3$ Solution 2: $(3, \infty)$

Intersection of Solution 1 and Solution 2: $[-4, \infty) \cap (3, \infty) = (3, \infty)$. Solution: $x \in (3, \infty)$

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Explanation of the solution:

Each problem involves solving one or more linear or rational inequalities.

1. Linear Inequalities: Simplify by distributing, combining like terms, and isolating $x$. Remember to reverse the inequality sign if multiplying or dividing by a negative number.
2. Rational Inequalities:
   • Move all terms to one side to get $\frac{P(x)}{Q(x)} \geq 0$ or $\frac{P(x)}{Q(x)} \leq 0$.
   • Find critical points by setting the numerator and denominator to zero.
   • Plot critical points on a number line, dividing it into intervals.
   • Test a value from each interval in the inequality to determine the sign.
   • The solution is the union of intervals that satisfy the inequality. Remember that values making the denominator zero are excluded.
3. Compound Inequalities (systems): Solve each inequality separately and then find the intersection of their solution sets.

Answer: (i) $x \in (-\infty, -1) \cup [1, \infty)$ (ii) $x \in (-\infty, -2) \cup [2, \infty)$ (iii) $x \in (-\infty, -1)$ (iv) $x \in (-12, -5)$ (v) $x \in [-16, -7)$ (vi) No solution ($\emptyset$) (vii) $x \in [8, \infty)$ (viii) $x \in (4, 7)$ (ix) $x \in (-\infty, 120)$ (x) $x \in (1/4, \infty)$ (xi) $x \in (-1, 3)$ (xii) $x \in (5, \infty)$ (xiii) $x \in (3, \infty)$