Inequality $7|x + 3| + 9 > 6$ Solve the absolute value inequality. Express the answer using interval notation. $$7|x + 3| + 9 > 6$$
My answer was $\left(-\frac{24}7, \infty\right)$ and $\left(-\infty,\frac{36}7\right)$. 
I multiplied the $7$ into the absolute value (following PEDMAS) and added the $9$. 
$7x + 21 +9 > 6$ is this the wrong way to solve? 
 A: $7|x+3|+9>6 \iff 7|x+3|>-3 \iff|x+3|>-\frac{3}{7}.$
But $|x+3| \geq 0$ by definition, so the set of all solutions is $\mathbb{R}$ (i.e. any real number is a solution).
(of course, if you allow complex $x$, then the set of all solutions is $\mathbb{C}$).
A: That method will not work. In problems like this, you want to isolate the absolute value expression on one side of the inequality sign. So in your problem, we get $7|x+3| > -3$, or $|x+3|>-\frac{3}{7}$. Since $|x+3|\ge 0$ for any value of $x$, this inequality is satisfied for any value of $x$, so your original problem has a solution set of $(-\infty,\infty)$.
Note that you can more simply just observe: well, $9>6$, and $7|x+3|$ is always nonnegative, so that this inequality always holds.
A: The absolute value of a number isn't always that number. So you can't treat it as just some type of parenthesis and thus you can't reduce $|x+3|$ to $x+3$. What you can do however is to use the following properties: 

If $X,a\in\mathbb R$ such that $|X|\geqslant a,$ then we have $$X\geqslant a{}{}{}\quad{}{}{}\text{or}{}{}{}\quad {}{}{}X{}{}{}\leqslant{}{}{}-{}{}{}a{}{}{}.{}{}{}$$
If $X,a\in\mathbb R$ such that $|X|\leqslant a,$ then we have $$X\leqslant a\quad\text{or}\quad X\geqslant-a.$$

Those facts can be intuitively understood by thinking of the absolute value of $X$ as the distance of $X$ from $0$. So if $|X|\geqslant a$ then that means that the distance of $X$ from $0$ is greater than $a$. By making a little drawing of an arbitrary number line, we can understand the content of the property: 

The same line of reasoning can be applied with the second property.
Those two properties are the fundamental ones at solving inequalities involving absolute values. You may use them in this case by first writing your original inequality as: $$\eqalign{
7|x + 3| + 9 \geqslant 6&\iff 7|x+3|\geqslant6-9
\\
&\iff 7|x+3|\geqslant-3 \\
&\iff |x+3|\geqslant-\tfrac37.}$$
Now letting $X=x+3$ and using the very first property we listed: $$X\geqslant-\tfrac37\quad\text{and}\quad X\leqslant-\tfrac37\,,$$ we get: $$x+3\geqslant-\tfrac37\quad\text{and}\quad x+3\leqslant-\tfrac37 \\ \Rightarrow \  x\geqslant-\tfrac{24}7\quad\text{and}\quad x\leqslant-\tfrac{24}7. \ \ \ \, $$ We can express those inequalities as $x\in\text{a set}$, to get: $$x\in\left[-\tfrac{24}7,+\infty\right) \quad\text{and}\quad x\in\left(-\infty,-\tfrac{24}7\right].$$ To finish things down, we need to determine when $x$ is an element of both sets. This can be obtained by taking the intersection of both sets (denoted $\cap$) to get that the set of solutions of our inequality is $(-\infty,\infty)$ which is just $\mathbb R.$
However, this way was long and time consuming. The shortcut that other answers point to is to write our inequality as: $$7|x+3|\geqslant-3\, ,$$ and to use the fact that the absolute value of any number is always positive. Therefore the set of solutions is $\mathbb R$.
