Find the sum of values of $x$ such that $|x+2| +|x-3| +|x+4| + |x+5| = 18$ I tried it by finding the different values at the $4$ inflection points of the graph. Then didn't know how to proceed. Am I correct till here?
 A: *

*$x \le -5 $ We need to solve $-(x+2)-(x-3)-(x+4)-(x+5)=18$ 

*$-5 \le x \le -4 $ We need to solve $-(x+2)-(x-3)-(x+4)+(x+5)=18$ 

*$-4 \le x \le -2$ We need to solve $-(x+2)-(x-3)+(x+4)+(x+5)=18$ 

*$-2 \le x \le 3$ We need to solve $(x+2)-(x-3)+(x+4)+(x+5)=18$ 

*$x \ge 3$ We need to solve $(x+2)+(x-3)+(x+4)+(x+5)=18$


Remember solve each of these and see if they are actually s solution to the original equation or even a contradiction with the inequality for that case equation. Hope this helps. 
A: Here is another solution, one that uses a higher-order theorem but lessens the work.
You can think of your problem as finding points on the number line where the sum of the distances to the points $-2,3,-4,$ and $-5$ is $18$. (This is because $|x+2|=|x-(-2)|$ is the distance from $x$ to $-2$, and so on.)
A well-known theorem is that the sum of the distances from a point $x$ to a given set of points is minimized at any median of the set of points and increases as $x$ moves away from the median. If the number of points in the set is even, as in your set of four numbers, any median between the two middle points give the minimum sum of distances. So the left-hand side of your equation is the smallest for any $-4\le x\le -2$ and increases as $x$ moves to the right away from $-2$ or to the left away from $-4$.
We easily see that the sum is $10$ for $-4\le x\le -2$, so there are exactly two values of $x$ that give the sum $18$. (There would be no solutions if we wanted a value smaller than $10$ and infinitely many solutions if we wanted the value $10$.) It takes only a small amount of work to see that the values are $x=-6.5$ to the left and $x=2$ to the right.
A: 
A graphical interpretation may be of help in eliminating some cases. All of the functions in the terms of the sum are just horizontal translations of the absolute value function. The sum of terms is too small in the interval $ \ -5 \ \le \ x \ \le  -2 \ $ to produce a value as large as 18, since the average value of the four terms needs to be $ \ \frac{18}{4} \ = \ 4.5 \ $ .  So we only need consider randomgirl's cases 1, 4, and 5 , of which the first two of these are the likeliest to offer a solution.  The set-up of those cases essentially comes from considering lines of slope 1 or -1 with appropriate $ \ y-$ intercepts.  We may also suspect that case 5 provides no solution (the sum of terms is too large, since the average of the largest three terms is greater than 6) and, indeed, the computed value for  $ \ x \ $ is not in the domain interval $ \ x \ > \ 3 \ $ .  So cases 1 and 4 offer the only possibilities (and the answer for case 1 can't be all that far below -5 ) .
A: For problems like this, it's handy to be able to quickly sketch graphs
of the sums and differences of functions.
In this case, we let $f(x) = |x+2| +|x-3| +|x+4| + |x+5|,$
which is the sum of four simple functions, 
$f_1(x) = |x+2|$, $f_2(x) = |x-3|$, $f_3(x) = |x+4|$, and $f_4(x) = |x+5|.$
All of these functions are shown in the plot below:

The four functions $f_1$, $f_2$, $f_3$, and $f_4$ are very easy to plot.
A key point that helps when plotting the sum of these four functions
is that all four functions are linear except at their points of inflection,
so if  you plot the sum of the functions at each of the four values
$x=-5$, $x=-4$, $x=-2$, and $x=3$, you can simply "connect the dots"
with straight segments to plot the function from $x=-5$ to $x=3$.
To the left of $x=-5$, all four functions have slope $-1$, so their sum
has slope $-4$; to the right of $x=3$, the sum has slope $4$.
That is all we need to know in order to completely plot the function
and find the two points where $f(x) = 18$.
Notice that between $x=-4$ and $x=-2$, we have two functions that have turned up
and have slope $1$, and two functions still going down with slope $-1$,
so the sum of the functions has slope zero,
that is, the graph of the function $f$ in this interval is a horizontal line.
That's an illustration of the more general theorem cited by
Rory Daulton: given a finite set of predetermined 
points on the real number line,
for any point $x$ in the region where there are the same number 
of predetermined points to the left as predetermined points to the right,
you will get the same sum of distances to all those points.
