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Solving Inequalities with Absolute Values

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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

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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