Multiple Absolute Value Inequalities/Equalities

Date: 08/09/2001 at 20:24:15
From: karen penner
Subject: Multiple absolute value inequalities/equalities

I can solve inequalities containing absolute values, but not if there 
are two in the same equation. Apparently there is some manipulation 
that I am unaware of.

Ex.  |x+7|<3 or |x-7|<3

I know that if I only have one set |x+7|>3, then I simply set up the 
equation as x+7<3 or x+7>-3 and solve. But I don't know what to do 
with both. The method I use to answer one absolute value inequality 
does not give me the right answer for an inequaltiy with two absolute 
values.

I have the same question with multiple absolute value equalities.
Ex. |2y+4|+|6-2y| = 22

What's the method?

Thank you.

Date: 08/09/2001 at 23:27:56
From: Doctor Peterson
Subject: Re: multiple absolute value inequalities/equalities

Hi, Karen.

I wish you had shown me your answer and your work, so I could see 
whether you are really wrong. Since you are able to work with x+7<3 or 
x+7>-3, which already consists of two inequalities, I don't see what 
would be different, except for a greater level of complexity. But I 
suspect your difficulty may be in handling "and" and "or" in these 
contexts, because your statement I just quoted is subtly wrong.

To solve

    |x+7|<3

you solve

    x+7 < 3    (for x >= -7)
and
    -(x+7) < 3 (for x <= -7)

I've said this a little differently than you: I included in each 
inequality the conditions under which it applies; |x+7| is equal to 
x+7 only when x >= -7. Now I can solve each case:

    x < -4  AND x >= -7
OR
    x > -10 AND x <= -7

If either of these cases is true, then the inequality is solved. These 
can be restated as compound inequalities,

    -7 <= x < -4
OR
    -10 < x <= -7

This says that any x that is EITHER between -7 and -4, OR between -10 
and -7, solves the problem. But these can be combined into one 
inequality, since together they cover the entire interval from -10 to 
-4:

    -10 < x < -4

This is the long way to solve this; we could have simply said that x 
is within 3 units of -7, since |x+7| is the distance from x to -7. But 
this method of breaking into cases and then combining them is useful 
in complicated situations.

Now to solve your whole problem, we just have to combine the results 
of the two parts:

    |x+7|<3 or |x-7|<3

The first, as we've just seen, comes out to

    -10 < x < -4

The second gives

    4 < x < 10

Putting these together, we have

    -10 < x < -4 OR 4 < x < 10

That is, again, any x that is EITHER between -10 and -4, OR between 4 
and 10, solves the problem, because it makes one or the other of the 
two inequalities true. On a number line, we can graph this as

      -10   -7    -4       0       4     7     10
    <--+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-->
       o===========o               o===========o

I'll just give a quick hint for your other problem, and let you write 
back for more if you have trouble.

    |2y+4|+|6-2y| = 22

Here you will need not two but three cases. Notice that

    |2y+4| = (2y+4)  if 2y+4 >= 0
            -(2y+4)  if 2y+4 <= 0

Since 2y+4 = 0 when y = -2, this means that the absolute value has to 
be treated differently depending on whether y is above or below -2. 
Similarly,

    |6-2y| = (6-2y)  if 6-2y >= 0
            -(6-2y)  if 6-2y <= 0

This will behave differently when y is above or below 3. So putting it 
together, the whole equation has to be considered when y < -2, when 
-2 <= y <= 3, and when y > 3. Write an equation in each of the three 
cases, and then put the results together as I did above.

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