Integers $a$ for which the equation $\big\lvert 6\lvert x\rvert -8\big\rvert = a+x$ has the most solutions $$\big\lvert 6\lvert x\rvert -8\big\rvert = a+x$$
I know this should be done graphically, looking at each case and seeing for which $a$ will it intersect the $x$-axis the most times, but I can't seem to get this one right. Graphing is tedious and I almost always end up with with five more or five integer solutions than I am supposed to.
Is there a way other than a graphical to do this? How? Also, a graphical actual solution would be nice.
 A: You can do it with no graphic like that:
For $x\leq-\dfrac 43$, then you have $-8-6x=a+x$ of potential solution $x=\dfrac{-a-8}{7}$. Since $x\leq-\dfrac 43$, you have $a\geq\dfrac 43$ as a condition
For $-\dfrac 43\leq x\leq0$ you have $8+6x=a+x$ of potential solution $x=\dfrac{a-8}{5}$. Since $-\dfrac 43\leq x\leq0$, you have $8\geq a\geq\dfrac 43$ as a condition
For $0\leq x\leq \dfrac 43$ you have $8-6x=a+x$ of potential solution $x=\dfrac{8-a}{7}$. Since $0\leq x\leq \dfrac 43$, you have $8\geq a\geq -\dfrac 43$ as a condition
For $\dfrac 43\leq x$ you have $6x-8=a+x$ of potential solution $x=\dfrac{a+8}{5}$. Since $\dfrac 43\leq x$, you have $a\geq -\dfrac 43$ as a condition
You combine everything and get $\dfrac 43\leq a\leq 8$ as the optimum range for which you will have four possible solutions. 
A: The standard way of dealing with absolute values is splitting the possibilities:
The furthest nested absolute value is $|x|$, which brakes at $x=0$, so:
Case 1: $x>0$. now, the expression becomes $|6x-8|=a+x$, where we have an absolute value which breaks at $6x=8$, meaning $x=\frac43$.
Case 1.1.: $x > \frac43$. The expression then becomes $6x-8=a+x$, which has one solution: $$\frac{a+8}{5}.$$
Case 1.2.: $x<\frac43$. The expression $8-6x = a+x$ again has only one solution, $$\frac{8-a}{7}.$$
Case 2: $x<0$. now, the expression is $|-6x-8| = a+x$, and the left side of the equation breaks at $x=-\frac43$.
Case 2.1.: $x>-\frac43$ The expression is $6x+8=a+x$ with a solution $$\frac{a-8}{5}$$
Case 2.2.: $x<-\frac43$. The expression is $-6x-8=a+x$ with a solution $$\frac{-8-a}{7}$$
A: Non-graphical solution has been treated already, hence I provide a plot, hope that might be of some help. 
Plotting $|6|x|-8|-x$ we obtain

hence choosing $a=2,3, ..., 7$ you obtain 4 solutions.
A: The left-hand function has turning points at $(-4/3,0),(0,8)$ and $(4/3,0)$.
$y=a+x$ meets those points at $a=4/3$, $a=8$ and $a=-4/3$
So if $a<-4/3$, it misses entirely, and so on.  Only four cases.
