Q.If $|x+2|\le9$, then $x$ belongs to

$|x+2|\le 9$ means $-9\le x+2\le 9$, which gives $x\in[-11,7]$.

By the definition of absolute value, $|x+2|\le 9$ is equivalent to $-9\le x+2\le 9$.

Subtracting 2 from all three parts: $-9-2\le x\le 9-2$, i.e. $-11\le x\le 7$.

So $x\in[-11,7]$.

Squaring the inequality (safe since both sides are $\geq0$) removes the modulus and reduces to a simple linear inequality.

We are given $|x-1|\geq|x-3|$. Since both sides are non-negative, we may square both sides without reversing the inequality:

$(x-1)^2\geq(x-3)^2$

Expanding, $x^2-2x+1\geq x^2-6x+9$.

The $x^2$ terms cancel: $-2x+1\geq-6x+9$.

Add $6x$ to both sides: $4x+1\geq9$, so $4x\geq8$, giving $x\geq2$.

So the solution set is $[2,\infty)$.

Solving $x^2-3|x|+2=0$ as a quadratic in $|x|$ gives two positive roots for $|x|$, each of which yields two real values of $x$, for 4 solutions total.

Let $y=|x|\ge 0$. Since $x^2=y^2$, the equation becomes:

$y^2 - 3y + 2 = 0 \implies (y-1)(y-2)=0 \implies y=1 \text{ or } y=2$

Both values are non-negative, so both are valid for $|x|$.

$|x|=1 \implies x=\pm1$ (2 solutions).

$|x|=2 \implies x=\pm2$ (2 solutions).

Total: 4 distinct real solutions.

Splitting the absolute value at $x=1$ and solving each case gives exactly two valid roots, $x=0$ and $x=1$.

Case $x\ge1$ (so $|x-1|=x-1$): $x^2+(x-1)=1 \Rightarrow x^2+x-2=0 \Rightarrow (x+2)(x-1)=0 \Rightarrow x=-2$ or $x=1$. Only $x=1$ satisfies $x\ge1$.

Case $x<1$ (so $|x-1|=1-x$): $x^2+(1-x)=1 \Rightarrow x^2-x=0 \Rightarrow x(x-1)=0 \Rightarrow x=0$ or $x=1$. Only $x=0$ satisfies $x<1$.

So the full solution set is $\{0,1\}$ — 2 solutions.

$|x+2|\le 8$ means $-8\le x+2\le 8$; subtracting 2 throughout gives $-10\le x\le 6$.

$|x+2|\le 8$ unfolds to the compound inequality $-8 \le x+2 \le 8$.

Subtract 2 from all three parts: $-8-2 \le x \le 8-2$, i.e. $-10 \le x \le 6$.

Since the original inequality is $\le$ (not strict), both endpoints are included, giving the closed interval $[-10,6]$.