Tutorial 2: More Algebra, Inequalities and Absolute values

CHECK THE LINK BELOW THIS HEADING FOR SOLUTIONS!
Remember to check the corresponding Questions on the memo only

Click here to view the solutions to the question below

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) \(\frac{3-2x}{1+2x} \leq 4\)

6. Solve \(3 \leq |x - 2| \leq 7\) for \(x\).



7. Solve for \(|x-3|^2 - 4|x - 3| = 12\) for \(x\). (Hint: First let \(z = |x -3|\))



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