How to solve equations having absolute values What is the procedure to solve equations that have multiple expression in absolute values?
Are there popular methods to solve them?
I recently learned, $$|x|≠ ± x $$
So I'm confused as to how does one solve equation that has multiple expression in absolute values like
$$|3x+4| +|2x-1| - \frac{1}{4} |x+5| = 5$$
– Thanks :)
 A: In the range $x<-5$ (note that when $x=-5, x+5=0$), we have 
$$|neg|+|neg|-\frac14|neg|=5$$ as all three expressions within the mod signs are negative in that range. 
Thus we can solve:
$$-(3x+4)-(2x-1)+\frac14(x+5)=5$$
and see that $x=-\frac{27}{19}\not<-5$ and therefore in this range there are no solutions.

In the range $-5<x<-\frac43$ (note that when $x=-\frac43, 3x+4=0$), we have $$|neg|+|neg|-\frac14|pos|=5$$
So we solve in the range $-5<x<-\frac43$: $$-(3x+4)-(2x-1)-\frac14(x+5)=5$$
and achieve $x=-\frac{37}{21}$ which IS in the correct range and therefore a solution.
Can you continue this?
A: First, note the definition: $$|f(x)| = f(x) \ \ \ \ \text{if   }\ \ \ \ f(x)\ge0\\|f(x)| = -f(x) \ \text{otherwise}$$
This, in very simple terms, means that any solution to $|f(x)| =$ something is either a solution to $f(x) =$ something, or a solution to  $-f(x) =$ something.
so, for instance, in your case, you have:
$$|f(x)|+|g(x)|-|h(x)|=5\\f(x)=3x+4\\g(x)=2x−1\\h(x)=\frac{x+5}{4}$$
A solution to this equation must also be a solution to one of these:
$$f(x)+g(x)+h(x)=5\\f(x)+g(x)−h(x)=5\\f(x)-g(x)+h(x)=5\\f(x)-g(x)−h(x)=5\\-f(x)+g(x)+h(x)=5\\-f(x)+g(x)-h(x)=5\\-f(x)-g(x)+h(x)=5\\-f(x)-g(x)−h(x)=5$$
It is simple enough to find all of the solutions to each of these, and then check to see if each of these is a solution to the original equation (they may not be, but every solution to the original will be a solution to at least one of these, so you wont miss any). This requires a fair amount of work, and you can be cleverer than this, but I find this to be the most intuitive approach.
To reiterate, any time I have |f(x)| in an equation, I can find the solutions to the equation with |f(x)| replaced with f(x) and the equation with |f(x)| replaced with -f(x), and simply check those values to see if they are solutions in the equation with |f(x)|. 

If you wish to delve a little bit deeper, you might notice that we do not have to check the solution against the original equation, but merely check that the solution lies in the set where the equations themselves are equal, this means solving the inequalities $$f(x) \ge 0 \\g(x)\ge 0 \\ h(x) \ge 0 \\f(x) \le 0 \\g(x)\le 0 \\ h(x) \le 0$$ we can then associate each of the set of 8 functions we created earlier with the set that satisfies the corresponding collection of inequalities. This would help us, for instance, in the case where, instead of solving $|f(x)|+|g(x)|-|h(x)|=5$ we were solving $|f(x)|+|g(x)|-|h(x)|>5$, where we have infinitely many solutions, and the only way to express them is as the intersection of the sets satisfying each condition separately. 
