Find all $x \in R$ that satisfy the inequality $|4x − 3| \le 11$ 
Find all $x \in R$ that satisfy the inequality $|4x − 3| \le 11$

good day, I am having a hard time solving this question I need a little help please.
 A: \begin{align}|4x-3|&\le 11\\-11\le4x-3&\le11\\-8\le4x&\le14\\-2\le x&\le\dfrac72\\x&\in\left[-2,\dfrac72\right]\end{align}
A: We have two cases to consider: if $4x-3<0$ and if $4x-3\geq 0$ (note that the case where $4x-3 = 0$ can be examined together with either $<0$ or $>0$, since $0$ is not positive or negative).
The 'pivot point' where $4x-3=0$ is at $x=3/4$. Thus, we can split our question into cases where $x<3/4$ and those where $x>3/4$. Let us do the former case:
Suppose $x<3/4$. Then $4x-3<0$ and therefore $|4x-3| = -(4x-3)$. Thus, we must solve the inequality $-(4x-3)\leq 11$. We have
\begin{align*}
-(4x-3)
&\leq 11
\\
-4x+3
&\leq 11
\\
-8
&\leq 4x
\\
-2
&\leq x.
\end{align*}
Great. However, we mustn't forget our original assumption: that $x<3/4$. Combining these two inequalities, we see that all $x\in [-2,3/4)$ satisfy our inequality.
I'll leave the $x>3/4$ case to you.
A: Case 1: When $4x-3\ge 0 \implies  x\ge 3/4$, the the eq. becomes $4x-3\le 11 \implies x\le 7/2$, the overlap of the two intervals is $[3/4,7/2]$.
Case 2: When $4x-3<0 \implies x<3/4$, now the eq. becomes $-(4x-3)\le 11 \implies x \ge -2$, the overlap of the two intervals is   $[-2,3/4)$.
The union of these two intervals gives $x \in [-2, 7/2]$, the final answer.
Also, $|4x-3\le 11 \implies -11\le (4x-3) \le 11 \implies  -8x \le 4x \le 14 \implies -2 \le x\le 7/2.$
A: If $0\leq a\leq b$ then $a^2\le b^2$, so you can square your inequality:
$$|4x-3|^2\leq 11^2$$
$$(4x-3)^2\leq 121$$
$$16x^2-24x+9\leq 121$$
$$16x^2-24x-112\leq 0\;\;\;/:8$$
$$2x^2-3x-14\leq 0$$
$$(2x-7)(x+2)\leq 0$$
$$...$$
