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Solving Inequalities with Absolute Values


Date: 09/16/98 at 23:55:09
From: Aamer Nazir
Subject: Solving inequalities with absolute values

Hi,

We are doing a chapter on absolute values and I have great difficulty 
in solving inequalities with absolute values. My instructor told me 2 
different ways of doing the same thing, and as a result I am confused. 
The problem is as follows:

    |X+1| <= 3X

One of the ways I am trying is breaking it apart. So:
 
   -(X+1) <= 3X (when X < 1)

and

      X+1 <= 3X  (when X >= -1)

First of all, I have a confusion about all this process. The reason is 
that I don't know how to put the solution on the number line as this 
method confuses me with the method while solving inequalities without 
absolute values. If you could just go on and write the solution for 
the inequality with the steps explained, it would be much appreciated. 
Also if there is another way of doing it I would like to know that 
too.

Thanks a lot in advance,
Aamer


Date: 09/17/98 at 12:49:10
From: Doctor Peterson
Subject: Re: Solving inequalities with absolute values

Hi, Aamer. Absolute values do make things more complicated, don't they? 

Let's go through this slowly. The problem is:

   |X + 1| <= 3X

As you said, we can break this into two cases. One solution will be 
obtained if we assume that X + 1 >= 0:

   (X + 1) <= 3X
         1 <= 2X   (subtracting X from both sides)
       1/2 <= X    (dividing by two)
         X >= 1/2  (reversing the order)

Now, we assumed that X + 1 >= 0, which means that:

   X >= -1

So the solution to this case requires both inequalities to be true:

   X >= 1/2 and X >= -1

If X >= 1/2, then it has to be >= -1, so we can drop the latter (one 
set includes the other), and we get:

   X >= 1/2

Now let's take the other case, assuming X + 1 < 0:

   -(X + 1) <= 3X
     -X - 1 <= 3X    (expanding the left side)
         -1 <= 4X    (adding X to both sides)
       -1/4 <= X     (dividing by 4)
          X >= -1/4  (reversing the order)

Now both this and the assumption have to be true for this to work:

   X >= -1/4 and X < -1

But if X >= -1/4, it can't be < -1, so this is impossible (the two sets 
have an empty intersection). So this case doesn't give us solutions.

Now we have to combine the two solutions, with an "or": either X 
satisfies the first solution, or X satisfies the second solution. Since 
the second is empty, we just have:

   X >= 1/2

as the solution.

Just for fun, let's graph the problem and see what's happening. I'll 
graph the left and right sides, and we'll see where the left is below 
the right:

                     |           / 3X
                     |          /
                     +         /
                     |         /
                     |        /
                     |       /
                     +       /
                     |      /      / |X + 1|
                     |     /     /
                     |     /   /
                     +    /  /
                     |    //
                     |   /
                     | //
     \               /  /
       \           / | /
         \       /   |/
           \   /     |/
  ---+-------*-------*-------+-------+-
               .    /|    \___________
                 .  /|      left side
                   / |      smaller
                  /  |
                  /  | .
                 /   |   .
     ^       ^   ^ ^ ^   ^   ^

You can see that the second case has no solutions because the absolute 
value turns the left side farther away from the right side. The -1/4 
we saw is where this would intersect if it continued in the other 
direction, as I have marked.

I don't know of a better way to solve this sort of problem. It's 
probably a good idea to check when you're done by trying some values 
in each domain you have seen while you worked on it: that is, at -1 
and -1/4 and 1/2, the three end points you have considered, and in 
each space between them, such as -2, -1/2, 0, and 1, where I put ^ in 
the graph. That helps me feel more confident - absolute values can 
shake the confidence of the best mathematician.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   


Date: 09/17/98 at 15:08:04
From: Doctor Anthony
Subject: Re: Solving inequalities with absolute values

Method (1).  Square both sides.

   x^2 + 2x + 1 <= 9x^2
              0 <= 8x^2 - 2x - 1
              0 <= (4x+1)(2x-1)

   or (4x+1)(2x-1) >= 0

Imagine the graph of y = 8x^2 - 2x - 1. This is a parabola cutting 
the x axis at -1/4 and 1/2.  Between these values the graph is below 
the x axis, so that y < 0  while we require y >= 0. So clearly y >= 0 
if x <= -1/4 or x >= 1/2.

Because we squared the equation one of these could be wrong, so test 
in the original inequality |x+1| <= 3x.  

Try x = -2. Is 1 <= -6? No! So we cannot accept the answer x <= -1/4.
Try x = 1. Is 2 <= 3? Yes! So we accept the answer x >= 1/2.

The solution is  x >= 1/2    
     
Method (2).  

   |x+1| <= 3x   

The critical point is where |x+1| passes through zero. This occurs 
when x = -1, so we consider two regions x <= -1 and x >= -1.

When x <= -1 write the equation:

   -(x+1) <= 3x
   -x - 1 <= 3x
      - 1 <= 4x
    - 1/4 <= x

This is not in the permitted region with x <=-1 so we do not accept 
this answer.

When x >= -1  write the equation:

   x+1 <= 3x
     1 <= 2x
   1/2 <= x

So x >= 1/2. This is in the permitted so we can accept this result.
The solution is x >= 1/2.
 
- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   


Date: 09/19/98 at 20:06:03
From: Aamer Nazir
Subject: Re: Solving inequalities with absolute values

Wow - thanks for the reply. This has been great help. I'll direct all 
my friends to your Web site for help.

Thanks again,
Aamer
    
Associated Topics:
High School Basic Algebra
Middle School Algebra

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