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

1. anonymous

By the way, b2 and 2b3 are exponents

2. anonymous

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

3. anonymous

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

4. anonymous

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: $|-5|=-(-5) \ \ \ \ whereas \ \ \ \ |3|=3$

5. anonymous

Make sense?

6. anonymous

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

7. anonymous

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

8. anonymous

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.

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

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

13. anonymous

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?

14. anonymous

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

15. anonymous

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

16. anonymous

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

17. anonymous

Ok so we have: $9|9-8x| = 2x+3$

18. anonymous

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

19. anonymous

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?

20. anonymous

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

21. anonymous

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

22. anonymous

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?

23. anonymous

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

24. anonymous

Are you talking about when 9-8x=0

25. anonymous

Because that wont satisfy the problem at a glance

26. anonymous

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

27. anonymous

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.

28. anonymous

AHHHH yes great point

29. anonymous

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)

30. anonymous

Yep! So first of all, we're supposed to divide 9 to both sides like this right? (9÷9)|9-8x|=2x+(3÷9)

31. anonymous

Yes you can start there

32. anonymous

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

33. anonymous

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

34. anonymous

ignore the double bar (typo)

35. anonymous

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

36. anonymous

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

37. anonymous

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

38. anonymous

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)): $a(b+c)=ab+ac \ \ \ or \ \ \ \ a(x+b)=ax+ab$

39. anonymous

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

40. anonymous

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

41. anonymous

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...

42. anonymous

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

43. anonymous

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

44. anonymous

So: $|-(2x-12)|=-(2x-12)$

45. anonymous

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

46. anonymous

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?

47. anonymous

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

48. anonymous

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

49. anonymous

Does this make sense?

50. anonymous

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

51. anonymous

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

52. anonymous

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

53. anonymous

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

54. anonymous

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

55. anonymous

Testing...

56. anonymous

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

57. anonymous

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

58. anonymous

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

59. anonymous

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

60. anonymous

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

61. anonymous

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

62. anonymous

Alright, I'll do that!

63. anonymous

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?

64. anonymous

For which case?

65. anonymous

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

66. anonymous

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

67. anonymous

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

68. anonymous

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

69. anonymous

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

70. anonymous

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

71. anonymous

Sorry I got hung up for a second

72. anonymous

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

73. anonymous

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

74. anonymous

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

75. anonymous

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

76. anonymous

77. anonymous

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

78. anonymous

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.

79. anonymous

*going to

80. anonymous

It is positive what do you mean?

81. anonymous

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

82. anonymous

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