Basic Mathematics

$3x^2 - 4|(x - 1)(x + 1)| + x - 1 = 0$
Sign scheme of $(x - 1)(x + 1)$: $+$ on $(-\infty, -1]$, $-$ on $[-1, 1]$, $+$ on $[1, \infty)$
Case-I: $x \in (-\infty, -1] \cup [1, \infty)$ (initial condition)
$\Rightarrow 3x^2 - 4(x^2 - 1) + x - 1 = 0$
$\Rightarrow x^2 - x - 3 = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{13}}{2}$: both $\frac{1 - \sqrt{13}}{2} \approx -1.30$ and $\frac{1 + \sqrt{13}}{2} \approx 2.30$ lie in the region (both accepted)
Case-II: $x \in (-1, 1)$ (initial condition)
$\Rightarrow 3x^2 + 4(x^2 - 1) + x - 1 = 0$
$\Rightarrow 7x^2 + x - 5 = 0$
$\Rightarrow x = \frac{-1 \pm \sqrt{141}}{14}$: both $\approx 0.78$ and $\approx -0.92$ lie in $(-1, 1)$ (both accepted)
$\Rightarrow$ Number of real roots = 4

Case 1: $3x - 2 = x$ gives $2x = 2$, so $x = 1$.
Case 2: $3x - 2 = -x$ gives $4x = 2$, so $x = 1/2$.
Both values satisfy $|3x - 2| = x$ because the right side $x$ is non-negative for each.
Hence $x = 1/2$ or $x = 1$.

For $x < -1$: $-(x+1) - (x-1) = -2x = 2$ gives $x = -1$, the boundary point.
For $-1 \le x \le 1$: $(x+1) + (1-x) = 2$, an identity, so every point of the interval works.
For $x > 1$: $(x+1) + (x-1) = 2x = 2$ gives $x = 1$, the boundary point.
Hence $x \in [-1, 1]$.

$||x-1| - 2| = 1$ gives $|x-1| - 2 = 1$ or $|x-1| - 2 = -1$.
So $|x-1| = 3$ or $|x-1| = 1$.
$|x-1| = 3$ gives $x = 4$ or $x = -2$.
$|x-1| = 1$ gives $x = 2$ or $x = 0$.
Hence the solution set is $\{-2, 0, 2, 4\}$.

For $x \le 1$: $(1-x)+(2-x)+(3-x) = 6 - 3x = 4$ gives $x = 2/3$, which lies in the region.
For $1 \le x \le 2$: $(x-1)+(2-x)+(3-x) = 4 - x = 4$ gives $x = 0$, not in the region, rejected.
For $2 \le x \le 3$: $(x-1)+(x-2)+(3-x) = x = 4$, not in the region, rejected.
For $x \ge 3$: $(x-1)+(x-2)+(x-3) = 3x - 6 = 4$ gives $x = 10/3$, which lies in the region.
Hence $x \in \{2/3, 10/3\}$.

$|x - 5| < 2$
$-2 < x - 5 < 2$
$-2 + 5 < x - 5 + 5 < 2 + 5$
$3 < x < 7$
The solution set consists of all real numbers that are greater than 3 and less than 7. The interval notation, for this solution set is $(3, 7)$. The graph of this solution set is an open segment from 3 to 7, each endpoint lying 2 units from 5.

$|x + 3| \geq 7$
$x + 3 \leq -7$ or $x + 3 \geq 7$
$x + 3 - 3 \leq -7 - 3$ or $x + 3 - 3 \geq 7 - 3$
$x \leq -10$ or $x \geq 4$
The solution set consists of all real numbers that are less than or equal to -10 or greater than or equal to 4. The interval notation for this solution set is $(-\infty, -10] \cup [4, \infty)$. The graphs of this solution set are two rays, each starting 7 units away from -3 on the number line.

$\left| \frac{3x + 3}{5} \right| < 3$
$-3 < \frac{3x + 3}{5} < 3$
$-15 < 3x + 3 < 15$
$-18 < 3x < 12$
$-6 < x < 4$
The solution set consists of all real numbers that are greater than -6 and less than 4. The interval notation for this solution set is $(-6, 4)$.

Let us write the given inequality in the form $x + 2\left|x - \frac{3}{2}\right| > |x + 1| - 1$ and mark on the number axis the points $x = -1$ and $x = \frac{3}{2}$ at which the expressions standing inside the modulus sign vanish. These points break the number axis into three intervals;
$x < -1$ (A), $-1 \leq x < \frac{3}{2}$ (B), $x \geq \frac{3}{2}$ (C).
Considering $x$ consecutively on each of these intervals, we obtain that the original inequality is equivalent to the set of the following three systems of inequalities:
System A: $x < -1$ and $x - 2\left(x - \frac{3}{2}\right) > -(x + 1) - 1$, which gives $3 > -2$, true for all $x < -1$.
System B: $-1 \leq x < \frac{3}{2}$ and $x - 2\left(x - \frac{3}{2}\right) > (x + 1) - 1$, which gives $-x + 3 > x$, i.e. $x < \frac{3}{2}$, so $-1 \leq x < \frac{3}{2}$.
System C: $x \geq \frac{3}{2}$ and $x + 2\left(x - \frac{3}{2}\right) > (x + 1) - 1$, which gives $3x - 3 > x$, i.e. $x > \frac{3}{2}$.
Taking the union of the three cases, the solution set is all real numbers except $x = \frac{3}{2}$, i.e. $\left(-\infty, \frac{3}{2}\right) \cup \left(\frac{3}{2}, \infty\right)$.

Case A: $x^2 + 3x \geq 0$ $\Rightarrow$ $x \leq -3$ and $x \geq 0$ ...(1)
We get the inequality $2x^2 + 3x - 2 \geq 0$ the solutions of which are $x \leq -2$ and $x \geq 1/2$.
We now have to choose from these solutions those which satisfy condition (1).
We get the solution for case A as $x \leq -3$ and $x \geq 1/2$.
Case B: $x^2 + 3x < 0$ $\Rightarrow$ $-3 < x < 0$ ...(2)
We get the inequality $-3x - 2 \geq 0$ $\Rightarrow$ $x \leq -2/3$.
We now have to choose from these solutions those which satisfy condition (2).
We get the solution for case B as $-3 < x \leq -2/3$.
Combining the solutions found in A and B, we get the answer: $x \leq -2/3$ or $x \geq 1/2$, i.e. $\left(-\infty, -\frac{2}{3}\right] \cup \left[\frac{1}{2}, \infty\right)$.

Mark on the number line those values of $x$ for which each of the expressions under the absolute value sign vanishes: $x = 1$, $x = 0$ and $x = -\frac{3}{2}$. These points break the number line into four intervals:
$x < -\frac{3}{2}$ (A), $-\frac{3}{2} \leq x < 0$ (B), $0 \leq x < 1$ (C), $x \geq 1$ (D).
Case A: the inequality becomes $(1 - x) + x - (2x + 3) > 3x + 4$, i.e. $-2x - 2 > 3x + 4$, giving $x < -\frac{6}{5}$; intersecting with $x < -\frac{3}{2}$ gives $x < -\frac{3}{2}$.
Case B: the inequality becomes $(1 - x) + x + (2x + 3) > 3x + 4$, i.e. $2x + 4 > 3x + 4$, giving $x < 0$; intersecting gives $-\frac{3}{2} \leq x < 0$.
Case C: the inequality becomes $(1 - x) - x + (2x + 3) > 3x + 4$, i.e. $4 > 3x + 4$, giving $x < 0$; intersecting with $0 \leq x < 1$ gives no solution.
Case D: the inequality becomes $(x - 1) - x + (2x + 3) > 3x + 4$, i.e. $2x + 2 > 3x + 4$, giving $x < -2$; intersecting with $x \geq 1$ gives no solution.
Combining the solutions found in A, B, C and D, we get the answer: $x < 0$, i.e. $(-\infty, 0)$.