# How to solve $| x - a | + | x + a | < b$ where $a \neq 0$?

The options are :

A ) no solution if $b\leq 2|a|$

B ) has a solution set ${ ( -b/2 , b/2 ) }$ if $b > 2|a|$

C ) has a solution set ${ ( -b/2 , b/2 ) }$ if $b < 2|a|$

D ) All of the above

My approach : ( I think its TOO long )

Case 1 : b = 2|a| => b > 0 and $b = \pm 2a$

Subcase A : a > 0

Sub-SubCase (i) $x \leq {-a}$

And so on :

So this leads to a total of : 15 subcases or so ....

My book has solved it using this graph , but I dont understand what it did : ( for example why did it consider |a| , -|a| , b/2 , -b/2 in x - axis )

(Note : Just a representation of figure , not the exact graph , not to scale )

Actually you only need three cases, not 16. Note the definition of $|x|$ is that its $x$ if $x\geq 0$ and $-x$ if $x<0$. Assume $a>0$ then:

Case 1: $x\geq a$. If this hold then $|x-a|=x-a$ and $|x+a|=x+a$. Thus we get $x-a+x+a<b$ i.e. $x<\frac{b}{2}$

Case 2: $-a\leq x< a$. If this hold then $|x-a|=-(x-a)= a-x$ and $|x+a|=x+a$. Thus we get $a-x+x+a<b$ i.e. $2a<b$.

Case 3: $x<-a$. If this hold then $|x-a|= -(x-a)=a-x$ and $|x+a| = -(x+a)=-x-a$. Thus we get $a-x-x-a<b$ i.e. $-2x <b$ which we may rewrite as $x>\frac{-b}{2}$.

Note that in case 1 and case 3 we have $x\geq a$ implies $\frac{b}{2}>x$ and $x<-a$ implies $\frac{-b}{2}<x$ correspondingly. Thus we get $a<\frac{b}{2}$ and $-a>\frac{-b}{2}$ which together imply that $a<\frac{b}{2}$ thus $2a<b$ has to hold. We conclude that if $2a<b$ then Case two will be true and in case $1$ and case $3$ $x<b/2$ and $x>-b/2$ respectively. Thus $x\in (-b/2,b/2)$. On the other hand if $2a>b$ then none of the cases will hold.

So the above only consider $a>0$. If $a<0$ then let $a_0=-a$ $\implies |x-a|=|x+a_0|$ and $|x+a| = |x-a_0|$. Thus the equation transforms from $|x-a|+|x+a|<b$ to $|x+a_0|+|x-a_0|<b$ which obviously is the same equation. Hence it has the same solutions, but this time we have $2a_0\geq b$ i.e. $-2a\geq b$. This together with our above solutions when $a>0$ mean the inequality have solution $(-b/2, b/2)$ if $2|a|\geq b$ hold. i.e. Case C).

Assume $a > 0$.

Geometrically, the distance from $x$ to $a$ plus the distance from $x$ to $-a$ cannot exceed $b$.

If $x$ is in between $-a$ and $a$, the sum of these distances is $2a$. So every $x \in [-a,a]$ is a solution if $2a < b$, and no point is a solution if $2a \ge b$.

If $x > a$, then $|x-a| + |x+a| = 2x$ so your solution is $2x < b$. Note this cannot be the case if $b \le 2a$.

If $x < -a$, then $|x-a| + |x+a| = -2x = |2x|$ so your solution is $2|x| < b$. Likewise, this is not the case if $b \le 2a$.