Chapter 5 - Linear Inequalities

Given $x > 0$, so $x+1 > 0$.
From $3 \le \frac{6}{x+1}$, multiplying by $x+1$ gives $3(x+1) \le 6 \implies x+1 \le 2 \implies x \le 1$.
From $\frac{4}{x+1} \le 3$, multiplying by $x+1$ gives $4 \le 3(x+1) \implies 4 \le 3x+3 \implies 1 \le 3x \implies x \ge \frac{1}{3}$.
Combining both, we get $\frac{1}{3} \le x \le 1$.

Let $|x-2| = y$. The inequality becomes $\frac{y-1}{y-2} \le 0$.
The critical points are $y=1$ and $y=2$. The expression is negative or zero when $1 \le y < 2$.
So, $1 \le |x-2| < 2$.
Case 1: $|x-2| \ge 1 \implies x-2 \ge 1$ or $x-2 \le -1 \implies x \ge 3$ or $x \le 1$.
Case 2: $|x-2| < 2 \implies -2 < x-2 < 2 \implies 0 < x < 4$.
Taking the intersection of both cases: $(0, 1] \cup [3, 4)$.

$\frac{1}{|x|-3} - \frac{1}{2} \le 0 \implies \frac{2 - (|x|-3)}{2(|x|-3)} \le 0 \implies \frac{5-|x|}{2(|x|-3)} \le 0$.
Let $t = |x|$. We need $\frac{5-t}{t-3} \le 0$.
Critical points are $t=3$ and $t=5$.
Using sign chart, the inequality holds for $t < 3$ or $t \ge 5$.
So $|x| < 3$ or $|x| \ge 5$.
$|x| < 3 \implies x \in (-3, 3)$.
$|x| \ge 5 \implies x \in (-\infty, -5] \cup [5, \infty)$.
Solution set: $(-\infty, -5] \cup (-3, 3) \cup [5, \infty)$.

$|x-1| \le 5 \implies -5 \le x-1 \le 5 \implies -4 \le x \le 6$.
$|x| \ge 2 \implies x \ge 2$ or $x \le -2$.
Intersection of $[-4, 6]$ and $(-\infty, -2] \cup [2, \infty)$ is $[-4, -2] \cup [2, 6]$.

Multiply by 4: $-20 \le 2-3x \le 36$.
Subtract 2: $-22 \le -3x \le 34$.
Divide by -3 (reverse inequality signs): $\frac{22}{3} \ge x \ge -\frac{34}{3}$.
So $x \in [-\frac{34}{3}, \frac{22}{3}]$.

First inequality: $4x - 2x \ge 17 - 3 \implies 2x \ge 14 \implies x \ge 7$.
Second inequality: $3x < 3 \implies x < 1$.
There is no value of $x$ that satisfies both $x \ge 7$ and $x < 1$.
Hence, no solution.

Profit $P(x) = R(x) - C(x)$.
For profit, $P(x) > 0$.
$43x - (26000 + 30x) > 0$.
$13x - 26000 > 0$.
$13x > 26000 \implies x > 2000$.
The company must sell more than 2000 cassettes.

Let $x$ be the third reading.
Average pH = $\frac{8.48 + 8.35 + x}{3} = \frac{16.83 + x}{3}$.
Condition: $8.2 < \frac{16.83 + x}{3} < 8.5$.
Multiply by 3: $24.6 < 16.83 + x < 25.5$.
Subtract 16.83: $7.77 < x < 8.67$.

Let $x$ be the litres of 3% solution added.
Total volume = $460 + x$.
Total acid content = $9\%$ of $460 + 3\%$ of $x = 0.09(460) + 0.03x = 41.4 + 0.03x$.
Condition: $5\% < \frac{41.4 + 0.03x}{460+x} < 7\%$.
$0.05 < \frac{41.4 + 0.03x}{460+x} < 0.07$.
First part: $0.05(460+x) < 41.4 + 0.03x \implies 23 + 0.05x < 41.4 + 0.03x \implies 0.02x < 18.4 \implies x < 920$.
Second part: $41.4 + 0.03x < 0.07(460+x) \implies 41.4 + 0.03x < 32.2 + 0.07x \implies 9.2 < 0.04x \implies x > 230$.
So, $230 < x < 920$.

Given $40 < C < 45$.
From $F = \frac{9}{5}C + 32$, we have $C = \frac{5}{9}(F - 32)$.
Substitute $C$: $40 < \frac{5}{9}(F - 32) < 45$.
Multiply by $\frac{9}{5}$: $40 \times \frac{9}{5} < F - 32 < 45 \times \frac{9}{5}$.
$72 < F - 32 < 81$.
Add 32: $104 < F < 113$.

Let the length of the shortest side be $x$ cm.
Then, longest side = $2x$ cm.
Third side = $(x + 2)$ cm.
Perimeter $P = x + 2x + (x + 2) = 4x + 2$.
Given $P > 166$.
$4x + 2 > 166 \implies 4x > 164 \implies x > 41$.
Thus, the minimum length of the shortest side is greater than 41 cm.

Given $155 < T < 205$.
Substitute $T$: $155 < 30 + 25(x - 3) < 205$.
Subtract 30: $125 < 25(x - 3) < 175$.
Divide by 25: $5 < x - 3 < 7$.
Add 3: $8 < x < 10$.
The depth will be between 8 km and 10 km.