Question

Solving absolute value Inequalities with more than one absolute value term:

Say for example, I have:
2|x - 4| < |2x - 4|

(this was a question I made up!)

How would I go about with solving this?

Answer 1

you have to break it down in to cases and it is a pain

Answer 2

I know how to do with one absolute value argument.

Answer 3

Ok wait so I will post, can you confirm if they are correct?

Answer 4

sure because the one you posted looks goofy

Answer 5

2|x - 4| < |2x - 4|


SHOOT Im stuck.

Answer 6

BEcause

Answer 7

WAit

Answer 8

ok we can do this

Answer 9

Could we work with how I used to do it with one argument? This is how it went.

Answer 10

Say I had the following:
|3 - 2x| ≤ 5

Answer 11

We would do it like this:
Case 1: 3 - 2x ≥ 0 AND 3 - 2x ≤ 5
or
Case 2: 3 - 2x < 0 AND -(3-2x) ≤ 5

Answer 12

So how would we go about when there are two arguments?

Answer 13

hold on these are different

Answer 14

and harder!

Answer 15

2|x - 4| < |2x - 4|
Ok so how would we do it, using the definition of absolute value?

Answer 16

\[|3-2x|<5\] easy algebra, we say
\[-5<2x-3<5\]
\[-8<2x<2\]
\[-4<x<1\]

Answer 17

Yea

Answer 18

but for this one, unless you want to graph (which may be easier) you have to break it in to cases

Answer 19

Now how are we suppose to solve this bad boy.

Answer 20

Ok let us go the case way.

Answer 21

ok
\[ |2x-4| = \left\{\begin{array}{rcc}
2x-4 & \text{if} & x \geq 2 \\
4-2x& \text{if} & x < 2
\end{array}
\right.
\]

Answer 22

How did you type that

Answer 23

and
\[2|x-4| = \left\{\begin{array}{rcc}
2x-8 & \text{if} & x \geq 4 \\
8-2x& \text{if} & x < 4
\end{array}
\right.
\]

Answer 24

trade secret

Answer 25

Wait, I am confused by how you got 4, and 2 for the "ifs"

Answer 26

oh that is easy enough,
\[|2x-4|=2x-4 \text{ if }2x-4>0\iff 2x>4\iff x>2\] etc

Answer 27

Got it.

Answer 28

ok so now we have to check over the intervals. suppose
\[x>4\] then you have
\[2x-8<2x-4\] which is true for all x, so we know this works if
\[x>4\]

Answer 29

So then by the definition of absolute value,
|x - 4| =
(x-4), if x ≥ 4
-(x-4), if x < 4

Answer 30

exactly

Answer 31

Ok now what are we suppose to do with these defined cases?

Answer 32

so we see that this inequality is satisfied for all x > 4 right? because if x is greater than 4 you are comparing
\[2x-8\] with
\[2x-4\] and it is always less

Answer 33

Give me one minute to go check above.

Answer 34

ok i wrote it out take your time

Answer 35

Wait how could you go x > 4? Is it not x ≥ 4?

Answer 36

ok fine be picky lol

Answer 37

You said we have to check the intervals, suppose "x > 4", but in the definition, we did x ≥ 4

Answer 38

Ok could you resay what you did with x ≥ 4?

Answer 39

sure and i will write it slowly

Answer 40

if
\[x\geq 4\] then
\[|2x-4|=2x-4\] and
\[2|x-4|=2x-8\] so the inequality is
\[2x-8\leq 2x-4\]
\[-8\leq-4\] which is always true. true for all x

Answer 41

now we have to check on the interval
\[2<x<4\]

Answer 42

So basically, you took these intervals and created a number line?

Answer 43

< -------------- 0 ---- 2 ---- 4 ------------------>

Answer 44

think i messed that up. on this interval
\[2|x-4|=8-2x\] and
\[|2x-4|=2x-4\] right?

Answer 45

yes exactly

Answer 46

so now we have to solve
\[8-2x<2x-4\]

Answer 47

Ok so you are saying we have to use the intervals that can be created from:
< -------------- 0 ---- 2 ---- 4 ------------------>
and test each one to see if they satisfy the inequalities

Answer 48

So in this case, we would have x < 2, 2 < x < 4, and x > 4

Answer 49

\[12<4x\]
\[3<x\] so now we know if works if x > 3

Answer 50

So in this case, we would have x < 2, 2 < x < 4, and x > 4
?

Answer 51

right you have to check three cases you have it!

Answer 52

YES!!!!!!!!!!!!!!!!!!!!!11

Answer 53

in the middle we see it works if
\[x>3\]

Answer 54

Ok so this is the basic idea: you get the definition, and you get the numbers for the number line after the "if" statement

Answer 55

Ok let me try

Answer 56

and we already know it works if
\[x>4\] so now we know it works if
\[x>3\]

Answer 57

go ahead.

Answer 58

Ok say we do x < 2,
then |x-4| will became -(x-4) and |2x -4| will became -(2x - 4)

2-(x-4) < -(2x - 4)
2(-x + 4) < (-2x + 4)
-2x + 8 < -2x + 4
8 < 4
This is not true.

Answer 59

right!

Answer 60

so putting it all together, the right hand piece always works, the middle tells you x > 3, and the left hand piece never works, so solution is
\[x>3\]

Answer 61

now this one was probably easier to graph because both of these are piecewise lines, and you could just graph them and see where one was above the other. but this is the algebra way to do it.

Answer 62

Would you mind showing me how you would graph it?

Answer 63

here is a picture, you will have to copy and past because the link won't work
http://www.wolframalpha.com/input/?i=2 |x-4|%3C|2x-4|

Answer 64

the absolute value signs mess up the link, so copy and paste rather than clicking

Answer 65

So the basic idea is this: you get the numbers that follow after the "if" in the definition, put them on the number line and establish intervals to test out with the original inequality. If it works out, then it is a solution?

Answer 66

So in this case we got 2, and 4, so if I put them on the number line, I would get:
x < 2, 2 < x < 4 and x > 4.

Answer 67

right. you have to use the correct definition of the function in each case

Answer 68

absolute value always a pain. it looks so easy when you first see it because it just says "make it positive" but it is a piecewise function so you constantly have to consider what interval you are working over.

Answer 69

you got this now right?

Answer 70

that didn't work

Answer 71

if you want to see how i wrote a piecewise function, right click on the expression and select "show source" it will give it to you