Question 50939
<pre><font size = 5><b>solve for |x+1| + |x-2|<u><</u> 2 
absolute value of x+1 and absolute value 
of x-2 is less than or equal to 2

Use the principle that

if A > 0 then |A| = A 
if A < 0 then |A| = -A
if A = 0 then |A| = 0

There are four cases to consider:

Case 1:  x+1 <u>></u> 0 AND x-2 <u>></u> 0
           x <u>></u> -1 AND x <u>></u> 2

This means x <u>></u> 2

|x+1| + |x-2| <u><</u> 2
    x+1 + x-2 <u><</u> 2
       2x - 1 <u><</u> 2
           2x <u><</u> 3
            x <u><</u> 3/2

This contradicts x <u>></u> 2, so Case 1 is impossible.

Case 2:  x+1 <u>></u> 0 AND x-2 <u><</u> 0
           x <u>></u> -1 AND x <u><</u> 2

This means -1 <u><</u> x <u><</u> 2

 |x+1| + |x-2| <u><</u> 2
  x+1 + -(x-2) <u><</u> 2
 x + 1 - x + 2 <u><</u> 2
             3 <u><</u> 2
  
This is never true, so Case 2 is impossible.

Case 3:  x+1 <u><</u> 0 AND x-2 <u>></u> 0
           x <u><</u> -1 AND x <u>></u> 2

This is impossible so Case 3 is impossible.

Case 4:  x+1 <u><</u> 0 AND x-2 <u><</u> 0
           x <u><</u> -1 AND x <u><</u> 2

This means x <u><</u> -1

   |x+1| + |x-2| <u><</u> 2
 -(x+1) + -(x-2) <u><</u> 2
     -x-1 - x+2  <u><</u> 2
           -2x+1 <u><</u> 2
             -2x <u><</u> 1
               x <u>></u> -1/2
     
But this contradicts x <u><</u> -1

So Case 4 is impossible also.

There is no solution.

Edwin</pre>