Question about absolute value in inequalities My book presents the following: $$7 \le x \le 9 $$ so $$ -1 \le x - 8 \le 1 $$ and $$ |x-8| \le 1$$
I usually get confused with the way that taking the absolute value of an expression works. Could anybody explain why the inequalities above are equivalent? I understand how the first and the second one are equivalent but not how the third one is equivalent to the second one. 
 A: On a number line, $|x-a|$ represents the (non-negative) distance between $x$ and $a$.  So $|x-8| \leq 1$ can be interpreted as: The distance between $x$ and $8$ is no more than $1$ unit.  This immediately gives you $7 \leq x \leq 9$.
To analyze this the other direction, if $a \leq x \leq b$, then think of the midpoint:  $\frac{a+b}{2}$, and half the distance from $a$ to $b$: $\frac{b-a}{2}$.  You obtain:
$$ \left| x - \frac{a+b}{2} \right| \leq \frac{b-a}{2}.$$
A: Suppose you had $|x| < a$ (note: the idea below holds the same for $\leq$ as well). Let's break this into two cases: $x \ge 0$ and $x < 0$.
If $x \ge 0$, then $|x| = x$ since $x$ is already nonnegative so what the inequality $|x| < a$ says in that case is that $x < a$.
If $x < 0$, then $|x| = -x$ since $x$ is negative. What the inequality then says is that $-x < a$, or equivalently that $x > -a$.
Putting these two pieces together we get that $-a < x < a$. In your case we have that $-1 \leq x - 8 \leq 1$. Well using the above expression, we have that this is equivalent to $|x-8| \leq 1$.
A: First of all, inequality relation is unaffected by 
$(i)$ the addition (hence, subtraction) of finite real numbers
$(ii)$ the multiplication (hence, division) by finite positive real numbers
Again if $y$ is real and $|y|\le b$ Clearly, $b\ge0$
$\implies y^2\le b^2\implies (y-b)\{(y-(-b)\}\le0$
$\implies$ either $y\ge b$ and  $y\le -b$ which is impossible as $b\ge0$
or $y\le b$ and $y\ge -b\implies -b\le y\le b$
