Question

How do you solve absolute values with negative signs? Like |-4b-8|+|-1-b2|+2b3=-2

Answer 1

By the way, b2 and 2b3 are exponents

Answer 2

When treating absolute values with variables in it is helpful to factor out as many negatives to remove them as possible. Example:
\[| -a +2b -c | = |-(a-2b+c)| = |a-2b+c|\]

Answer 3

Yep. I get really confused whenever variables are thrown in absolute values, since I've seen them solved so many different ways...

Answer 4

The reason this works is because if the argument of the absolute value (the quantity between the bars) is negative then the absolute value returns the NEGATIVE of that:
\[|-5|=-(-5) \ \ \ \ whereas \ \ \ \ |3|=3\]

Answer 5

Make sense?

Answer 6

Um... oh, yeah! That does kind of make sense! I'm kind of confused on what you showed me before this reply though...

Answer 7

So looking at the reposting of the problem (thank you @zepdrix, didnt realize those were powers at first)... You will have two cases to treat

Answer 8

Assuming that b is a real number... its square will always be positive so factor out the negative like I showed in my generic example and get rid of it.

Answer 9

That term you can then drop the absolute value bars since it will be strictly positive

Answer 10

Oh well I didn't want to jump right at the answer I wanted to show you an example of some generic statement with absolute value bars and show the process of factoring out a negative

Answer 11

Then my second post was to show you why I could simply eliminate it after it was factored

Answer 12

Can we actually start with this equation? I still have trouble understanding stuff like this, and what you're going into is really giving me a headache:
9|9-8x|=2x+3

Answer 13

Hahaha well thats good... think about it like lifting weights... If you lifted a feather would your arm muscles burn during the motion? Would they ache the next day?

Answer 14

Only with harder problems do you build your mental muscles to get stronger :D... sure no problem lets start there

Answer 15

Yeah, I know... I really need to wrap my mind around the simple stuff first though! I get confused so easily! D:

Answer 16

Its ok dont worry about it we all have to go through it at some point :D:D

Answer 17

Ok so we have:
\[9|9-8x| = 2x+3\]

Answer 18

And you divide 9 to both sides right? I got a fraction, so I wasn't sure of myself there either.

Answer 19

This type of problem will have two cases:
First case: When (9-8x) < 0
Second Case: When (9-8x) > 0
Think you can guess why we will need to have two cases?

Answer 20

Yes eventually but lets deal with the hard bit first.... then after that it should be easy yes?

Answer 21

Okay... you need two cases to show both possible solutions to the absolute value right?

Answer 22

Exactley... AND we can exploit the two properties I initially posted to completely remove the absolute value bars from the problem.... once we have split it up into two separate equations subject to those conditions... make sense?

Answer 23

And eventually you'll need to check for extraneous solutions too right?

Answer 24

Are you talking about when 9-8x=0

Answer 25

Because that wont satisfy the problem at a glance

Answer 26

Well lets first handle the problem then you can refresh my memory about extraneous solutions... its seems to be blank spot for me right now

Answer 27

Sorry I didn't see you send out that reply. But yeah, I never thought the absolute value signs could be changed into parentheses...

The question asks me to check for extraneous solutions, which is when one of the solutions turns out to not satisfy the solution.

Answer 28

AHHHH yes great point

Answer 29

Once you set up the two equations you solve for them... BUT you should ALWAYS check to make sure your solutions satisfy the original system (equation)

Answer 30

Yep! So first of all, we're supposed to divide 9 to both sides like this right?

(9÷9)|9-8x|=2x+(3÷9)

Answer 31

Yes you can start there

Answer 32

And I did it right? It's correct to get a fraction, right?

Answer 33

So we have
\[9|9-8x| = 2x+3 \\ \rightarrow \ \ \|9-8x| = \frac{2}{9}x+\frac{1}{3} \]

Answer 34

ignore the double bar (typo)

Answer 35

So we have
\[9|9-8x| = 2x+3 \\ \rightarrow \ \ \ |9-8x| = \frac{2}{9}x+\frac{1}{3} \]

Answer 36

What?? You can divide 9 into 2x too? Man, I never thought you could do that...

Answer 37

Okay, so once it becomes |9-8x|=2/9x+1/3
please show me how you set up the 2 possible solutions. I always get confused

Answer 38

Oh yea sure whenever you divide (or multiply) a number to a whole side it DISTRIBUTES to sums. For any real numbers (or variables that take on real numbers (x)):
\[ a(b+c)=ab+ac \ \ \ or \ \ \ \ a(x+b)=ax+ab\]

Answer 39

So just let a be 1/9 and your'e golden :D

Answer 40

Ok I will do an example problem and then I want you to try your'e problem:

Answer 41

I just always get confused when there's a negative number within the absolute value. Please help me figure out what to do when those show up...

Answer 42

For: \[ |2x-12|\] If
\[ 2x-12<0\] Then
\[-(2x-12)>0\] ALWAYS
Please take a second to consider this

Answer 43

If
\[-(2x-12)\] is always positive then the absolute value bars just return the quantity. I.e. if:
\[a>0 \ \ \ \ \rightarrow \ \ \ \ |a|=a\]

Answer 44

So:
\[|-(2x-12)|=-(2x-12)\]

Answer 45

Now I have given you the seeds to actually solve both parts... so please do that and post them here when you are done :D

Answer 46

So which side of the equation would you normally do that to? I thought you were supposed to do it to the right side, right?

Answer 47

Forget sides... we are just talking about the absolute valued term... the goal here is to figure out what to "substitute" in for the absolute value bars (in each specific case) so they are no longer in the original problem

Answer 48

Think about it as breaking the problem up into the original and a new sub problem.... once we figure out the sub problem (for both cases in the original problem) we will know what will take the place of the absolute value term

Answer 49

Does this make sense?

Answer 50

Alright... I'm still a little confused, but I'll try and solve it now

Answer 51

Bingo... thats the spirit... sometimes it just takes the courage to plow ahead and see what happens :D

Answer 52

Whats the worst that can happen... you get it wrong? Well at least now you will definitely know one thing that DOESNT work :D be optimistic :D:D

Answer 53

Alright, so when you first start off with |9-8x|=2/9x+1/3
|9-8x| becomes -(9-8x)? On both sides?

Answer 54

No only substitute in for the absolute value term... btw if you dont know how to use the equations tab could you please enclose your fractions in () like (2/9) for ease of reading

Answer 55

Testing...

Answer 56

Alright, I will. Sooo... only do -(9-8x) on the positive side?

Answer 57

OK whew.... If I randomly stop talking its because open study has locked up on me AGAIN today

Answer 58

Hopefully this doesnt happen but be warned open study has not cooperated with me today

Answer 59

I'm so sorry for my confusion. Math is definitely not my easiest class...

Answer 60

Oh yes sorry... try and be precise with you language... sooo you are going to substitute in -(9-8x) in for the |9-8x| for the case when 9-8x<0

Answer 61

Gotta work it then... Just like muscles... If you dont work them they will never grow and stay weak

Answer 62

Alright, I'll do that!

Answer 63

So once I do that, I got:
-9+8x=(2/9)x+(1/3) or -9+8x= -((2/9)x+(1/3))

Correct or incorrect?

Answer 64

For which case?

Answer 65

|9-8x|=2/9x+1/3

Answer 66

Sorry I forgot to type it with parentheses:
|9-8x|=(2/9)x+(1/3)

Answer 67

Which case though be specific as to what you are trying to say... remember the two cases? Which case belongs with which substitution

Answer 68

I hate to be pedantic, but this is a very important distinction because your answer may or may not be right depending on which case you assign them to... keep that in mind

Answer 69

I don't know honestly. What do you mean by that?

Answer 70

I'm ready to give up at this point... this is getting too complicated! :(

Answer 71

Sorry I got hung up for a second

Answer 72

Ok watch how I state 1 of the cases... and note the goal of math isnt just to get the answer, but to express that answer and your work in a way that someone reading it can know what you are doing

Answer 73

When 9-8x<0 ==> -(9-8x)>0
Thus |9-8x|=|-(9-8x)|=9-8x
Then you sub into your problem and solve for x below

Answer 74

That symbol btw ==> is just my way of doing an arrow... you read it as "implies"

Answer 75

But when you have -(9-8x) aren't you supposed to distribute the negative sign to both numbers, thus making it 9+8x?

Answer 76

AHHHH Im soooo sorry I made a typoe please hold

Answer 77

When 9-8x<0 ==> -(9-8x)>0
Thus |9-8x|=|-(9-8x)|=-(9-8x)=8x-9
Then you sub into your problem and solve for x below

Answer 78

I still don't understand why it doesn't become positive... thank you so much for your help, but I think I'm just goint to try and get help from my teacher.

Answer 79

*going to

Answer 80

It is positive what do you mean?

Answer 81

I said 9-8x< SOOO [-(9-8x)]>0... Its the whole quantity in the square brakets that are positive here so the whole quantity is unaffected by the absolute value signs since it is positive

Answer 82

If you prefer distribute the minus sign first then you will have it correct... btw you did that when you noticed my error only you didnt distribute correctly... check my corrected post a couple above