Tutorial 2: More Algebra, Inequalities and Absolute values
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1. Solve the equation
(a) \(\frac{1}{t-1} + \frac{t}{3} = \frac{1}{3}\)
(b) \(\frac{1}{x} = \frac{4}{3x} + 1\)
(c) \(\frac{z}{5} = \frac{3}{10}z + 7\)
(d) \(r-2[1-3(2r+4)] = 61\)
(e) \((t-4)^{2} = (t+4)^{2} + 32\)
(f) \(\frac{2}{3}x - \frac{1}{4} = \frac{1}{6}x - \frac{1}{9}\)
(g) \(\frac{4}{x-1} + \frac{2}{x + 1} = \frac{35}{x^2 - 1}\)
(h) \((\frac{1}{x-1} 2 \frac{2}{x^2}\)
(i) \(\sqrt{2x + 1} + 1 = x\)
(j) \(\sqrt{\sqrt{x-5}+x} = 5\)
(k) \(4(x + 1)^{\frac{1}{2}} - 5(x + 1)^{\frac{3}{2}} + (x + 1)^{\frac{5}{2}} = 0\)
2. Solve the inequality. Express the solution in interval form and illustrate the solution set on the real number line.
(a) \(4-3x \leq -(1+8x)\)
(b) \( \frac{2x+1}{x-5} \leq 3\)
(c) \( \frac{(x-1)^2}{(x+1)(x+3)}\)
(d) \(\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x + 2}\)
3. Solve for x:
(a) \(|7x + 3| = 17\)
(b) \(|3x + 7| - \frac{3}{4} = 0\)
(c) \(|9x| -11 = x\)
(d) \(2x-7 = |x + 1|\)
(e) \(|4x + 5| = |8x - 3|\)
(f) \(|\frac{x-3}{x+4}| = 5\)
4. Solve for x and express the solution in terms of intervals.
((a) \(|5 - 2x| \geq 4\)
(b) \(|7 - x| \leq 5\)
(c) \(\frac{1}{|3x + 1|} \geq 5\)
(d) \(|\frac{1}{2}x - 1| \geq 2\)
(e) \(\frac{2}{x + 3} < 1 \)
5. Solve for x and express the solution in terms of intervals. Use the fact that \(|a|<|b|\) or (\(\leq\)) if and only if \( a^2 < b^2 \) or (\(\leq\))
(a) \(\frac{1}{|x-3|} - \frac{1}{x + 4} \geq\)
(b) \(|3x| \leq |2x - 5|\)
(c) \(|2x + 1| > |x - 5|\)
(d) \(\frac{3-2x}{1+2x} \leq 4\)
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