Modulus Equations $$ |x + 1| + |x − 1| = x + 4$$
The only way I can solve this equation is to graph it...Through graphing, I get the following solutions:
$$x = -\frac{4}{3}, 4$$
Is their a general algebraic method which could be used to solve this and others of this type without having to graph them?
Any help will be much appreciated, thanks in advance.
 A: Casework is an annoying but good way to deal with relatively simple equations with absolute value signs. Here, you would need to consider three cases: $x\leq -1$, $-1 < x < 1$ and $x \geq 1$.
For example $x \geq 1$ yields $|x-1| = x-1$ and $|x+1| = x+1$, so your equation becomes $2x = x+4$ and thus $x=4$. 
The other two cases are analog.
A: Notice that either $|x|=x$, either $|x|=-x$ depending on the sign of $x$. Thus, by examining a finite number of cases, you can eliminate the absolute values.


*

*If $x\leq-1$, both numbers in the absolute values are negative (or zero), your equation is thus $(-1-x)+(1-x)=x+4$.

*If $-1<x<1$, the first is positive while the second is negative, and your equation is $(x+1)+(1-x)=x+4$.

*If $x\geq1$, both are positive or zero and the equation is $(x+1)+(x-1)=x+4$.
In the first case, you get $3x=-4$ or $x=-4/3$, which is $\leq-1$, so it's a solution.
In the second case, $x=-2$, but it's not between $-1$ and $1$, hence it's not a solution.
In the last case, $x=4$, is $\geq1$ so it's also a solution.
Notice that in the second case, removing the absolute values has introduced a "false" solution. That's why you have to check that the $x$ you find lies in the expected interval.

More details about removing the absolute value. For $|x+1|$ for instance, you know that $x+1\geq0$ for $x\geq-1$, hence $|x+1|=x+1$ for $x\geq -1$. Likewise, $|x+1|=-(x+1)$ for $x<-1$, since then $x+1<0$.
The same method apply to $|x-1|$, where you have to test the cases $x\geq1$ and $x<1$.
All in all, you have three intervals: $x\leq-1$ and $x\geq1$ where numbers inside the absolute values are either both negative or both positive, and $-1<x<1$ where one is positive, but not the other.
In each case, you replace each $|x+1|$ by $x+1$ or $-(x+1)$ and $|x-1|$ by $x-1$ or $-(x-1)$, according to the interval where $x$ lies.
