1) Set operations with \(U = \{1,2,3,\dots,11\}\), \(A = \{1,3,5,7,11\}\), \(B = \{3,5,7,9\}\): 1. Find complements: \(A' = U \setminus A = \{2,4,6,8,9,10\}\) \(B' = U \setminus B = \{1,2,4,6,8,10,11\}\) 2. Compute: \(\text{a) } A' \cap B' = \{2,4,6,8,10\}\) (elements in both complements) \(\text{b) } A' \cup B' = \{1,2,4,6,8,9,10,11\}\) (elements in either complement) \(\text{c) } (A \cap B)' = U \setminus (A \cap B)\), first \(A \cap B = \{3,5,7\}\), so complement is \(\{1,2,4,6,8,9,10,11\}\) \(\text{d) } (A \cup B)' = U \setminus (A \cup B)\), first \(A \cup B = \{1,3,5,7,9,11\}\), so complement is \(\{2,4,6,8,10\}\) 2) Prime number universe \(U = \{2,3,5,7,11,13,17,19,23,29,31,\dots\}\) Sets defined by inequalities: \(P = \{x:3(x+1) \geq 2(x+10)\}\), solve inequality: \(3x+3 \geq 2x+20 \implies x \geq 17\) So \(P = \{x \in U: x \geq 17\}\) \(Q = \{x \in U: 7 < x < 31\} = \{11,13,17,19,23,29\}\) Then complements: \(P' = \{p \in U : p < 17\} = \{2,3,5,7,11,13\} \) \(Q' = \{p \in U: p \leq 7 \text{ or } p \geq 31\} = \{2,3,5,7,31, \dots\}\) Compute: \(\text{a) } (P \cup Q)' = U \setminus (P \cup Q)\) \(P \cup Q = \{p \in U: p \geq 11\}\) since \(Q\) includes primes from 11 to 29 and \(P\) all primes \(\geq 17\) So \((P \cup Q)' = \{2,3,5,7\}\) \(\text{b) } P' \cap Q' = \{p < 17\} \cap \{p \leq 7 \text{ or } p \geq 31\} = \{2,3,5,7\}\) \(\text{c) } (P \cap Q)' = U \setminus (P \cap Q)\) \(P \cap Q = \{p: x \geq 17 \} \cap \{7 < x < 31\} = \{17,19,23,29\}\) So \((P \cap Q)' = U \setminus \{17,19,23,29\} = \text{all primes except } 17,19,23,29\) 3) Solve inequalities: 3a) \(|\frac{x}{3}+7| + 5 > 6\) Step 1: Subtract 5 \(|\frac{x}{3}+7| > 1\) Step 2: Absolute value inequality means \(\frac{x}{3}+7 > 1 \quad \text{or} \quad \frac{x}{3}+7 < -1\) Solving first: \(\frac{x}{3} > -6 \Rightarrow x > -18\) Second: \(\frac{x}{3} < -8 \Rightarrow x < -24\) Final solution: \(x > -18 \quad \text{or} \quad x < -24\) Interval notation: \(( -\infty, -24 ) \cup (-18, \infty )\) 3b) \(\frac{2}{3}(x+3) \leq \frac{2}{6}(x-4) + \frac{x}{2} + 1\) Multiply terms: \(\frac{2}{3}x + 2 \leq \frac{1}{3}x - \frac{4}{3} + \frac{x}{2} + 1\) Combine right side: \(\frac{1}{3}x + \frac{x}{2} + 1 - \frac{4}{3} = \frac{2}{6}x + \frac{3}{6}x + \frac{3}{3} - \frac{4}{3} = \frac{5}{6}x - \frac{1}{3}\) Inequality: \(\frac{2}{3}x + 2 \leq \frac{5}{6}x - \frac{1}{3}\) Multiply by 6 to clear denominators: \(4x + 12 \leq 5x - 2\) Bring terms to one side: \(12 + 2 \leq 5x - 4x \Rightarrow 14 \leq x \Rightarrow x \geq 14\) 3c) Compound inequality: \(\frac{9}{4} \leq \frac{1}{2}(3x+2) - \frac{3}{4}(2x-3) < \frac{13}{4}\) Simplify the expression inside: \(\frac{1}{2}(3x+2) = \frac{3}{2}x + 1\) \(\frac{3}{4}(2x-3) = \frac{3}{4}*2x - \frac{3}{4}*3 = \frac{3}{2}x - \frac{9}{4}\) Difference: \(\frac{3}{2}x + 1 - (\frac{3}{2}x - \frac{9}{4}) = 1 + \frac{9}{4} = \frac{13}{4}\) So expression simplifies to \(\frac{13}{4}\) Check inequality: \(\frac{9}{4} \leq \frac{13}{4} < \frac{13}{4}\) Since \(\frac{13}{4} < \frac{13}{4}\) is false, no \(x\) satisfies the strict inequality. So no \(x\). 3d) \(-3 < \frac{2x+5}{3} \leq 5\) Multiply all parts by 3: \(-9 < 2x + 5 \leq 15\) Subtract 5: \(-14 < 2x \leq 10\) Divide by 2: \(-7 < x \leq 5\) Interval notation: \((-7, 5]\) 4) Express inequalities as absolute value inequalities: 4a) \(-3 < x < 11\) Center is midpoint: \(\frac{-3 + 11}{2} = 4\) Distance from center to boundary: \(11 - 4 = 7\) Absolute value form: \(|x - 4| < 7\) 4b) \(-\frac{7}{3} \leq x \leq 1\) Midpoint: \(\frac{-7/3 + 1}{2} = \frac{-7/3 + 3/3}{2} = \frac{-4/3}{2} = -\frac{2}{3}\) Distance: \(1 - (-2/3) = 1 + 2/3 = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}\) Absolute value form: \(|x + \frac{2}{3}| \leq \frac{5}{3}\) 4c) \(3 < x < \frac{9}{2}\) Midpoint: \(\frac{3 + \frac{9}{2}}{2} = \frac{3 + 4.5}{2} = \frac{7.5}{2} = 3.75 = \frac{15}{4}\) Distance: \(\frac{9}{2} - 3.75 = 4.5 - 3.75 = 0.75 = \frac{3}{4}\) Absolute value form: \(|x - \frac{15}{4}| < \frac{3}{4}\)