Solving Linear Inqualities

Date: 06/10/99 at 16:55:32
From: leon
Subject: Linear inqualities

How can I solve: 9 less than or equal to 2x-1 less than or equal to 
15 for x?

Date: 06/11/99 at 18:17:49
From: Doctor Rick
Subject: Re: Linear inqualities

Hi, Leon.

One way to think about this is to picture a number line.

                      2x-1
            |<=====================>|
 ---+---+---+---+---+---+---+---+---+---+---+---
    7   8   9   10  11  12  13  14  15  16  17

If you add 1 to 2x-1, you need to add 1 to every number on the number 
line:

                       2x
            |<=====================>|
 ---+---+---+---+---+---+---+---+---+---+---+---
    8   9   10  11  12  13  14  15  16  17  18

If you divide 2x by 2, you need to divide every number by 2:

                        x
            |<=====================>|
 ---+---+---+---+---+---+---+---+---+---+---+---
    4       5       6       7       8       9

And there is your answer: 5 <= x <= 8.

You could have gotten it by using the same rules you use with 
equations. You really have two inequalities:

  9 <= 2x-1    and    2x-1 <= 15

I'll solve both of these at the same time (as I did using the number 
line):

  10 <= 2x     and    2x <= 16    (add 1 to all 4 sides)

   5 <= x      and     x <= 8     (divide all 4 sides by 2)

Now reassemble the inequalities into one compound inequality:

   5 <= x <= 8

I get the same answer. 

I showed you the number-line way so you can see the one big difference 
between working with equations and working with inequalities. Let's 
solve this inequality:

  9 <= 1-2x <= 15

                      1-2x
            |<=====================>|
 ---+---+---+---+---+---+---+---+---+---+---+---
    7   8   9   10  11  12  13  14  15  16  17

If you subtract 1 from 1-2x, you need to subtract 1 from every number 
on the number line:

                       -2x
            |<=====================>|
 ---+---+---+---+---+---+---+---+---+---+---+---
    6   7   8   9   10  11  12  13  14  15  16

If you divide -2x by -2, you need to divide every number by -2:

                        x
            |<=====================>|
 ---+---+---+---+---+---+---+---+---+---+---+---
   -3      -4      -5      -6      -7      -8

Whoa, something's wrong - the number line doesn't look right. Let's 
walk around to the back side and look at it again:

                        x
            |<=====================>|
 ---+---+---+---+---+---+---+---+---+---+---+---
   -8      -7      -6      -5      -4      -3

That's better. Now smaller numbers are on the left. The inequality is:

  -7 <= x <= -4

Now I'll go through this inequality using the normal equation-solving 
technique:

  9 <= 1-2x    and    1-2x <= 15

  8 <= -2x     and    -2x <= 14    (subtract 1 from all 4 sides)

  -4 <= x      and     x <= -7     (divide all 4 sides by -2)

What does this look like on the number line? x must be in this region:

                                    |<===========
 ---+---+---+---+---+---+---+---+---+---+---+---
   -8      -7      -6      -5      -4      -3

AND  in this region:

===========>|
 ---+---+---+---+---+---+---+---+---+---+---+---
   -8      -7      -6      -5      -4      -3

There is no number that is in both regions. This is the wrong answer, 
but where did we go wrong? Remember how we had to go around to the 
back of the number line, how we had to turn it around when we divided 
by -2? When you do that, you have to reverse the direction of each 
inequality:

  -4 >= x      and     x >= -7

Putting these inequalities together, we have

  -4 >= x >= -7

or, writing smaller numbers on the left as we usually do,

  -7 <= x <= -4

That's the correct answer.

So here is the one new trick in working with inequalities: you must be 
careful when multiplying or dividing. If you multiply or divide by a 
negative number, the inequality signs must be turned around. This 
wasn't needed in the problem you presented, but I couldn't honestly 
leave it out when showing you how to solve inequalities.

I hope this has been of some help.

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