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dinorap1

  • one year ago

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

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  1. dinorap1
    • one year ago
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    By the way, b2 and 2b3 are exponents

  2. anonymous
    • one year ago
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    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. dinorap1
    • one year ago
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    Yep. I get really confused whenever variables are thrown in absolute values, since I've seen them solved so many different ways...

  4. anonymous
    • one year ago
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    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
    • one year ago
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    Make sense?

  6. dinorap1
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    That term you can then drop the absolute value bars since it will be strictly positive

  10. anonymous
    • one year ago
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    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
    • one year ago
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    Then my second post was to show you why I could simply eliminate it after it was factored

  12. dinorap1
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Only with harder problems do you build your mental muscles to get stronger :D... sure no problem lets start there

  15. dinorap1
    • one year ago
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    Yeah, I know... I really need to wrap my mind around the simple stuff first though! I get confused so easily! D:

  16. anonymous
    • one year ago
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    Its ok dont worry about it we all have to go through it at some point :D:D

  17. anonymous
    • one year ago
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    Ok so we have: \[9|9-8x| = 2x+3\]

  18. dinorap1
    • one year ago
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    And you divide 9 to both sides right? I got a fraction, so I wasn't sure of myself there either.

  19. anonymous
    • one year ago
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    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
    • one year ago
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    Yes eventually but lets deal with the hard bit first.... then after that it should be easy yes?

  21. dinorap1
    • one year ago
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    Okay... you need two cases to show both possible solutions to the absolute value right?

  22. anonymous
    • one year ago
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    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. dinorap1
    • one year ago
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    And eventually you'll need to check for extraneous solutions too right?

  24. anonymous
    • one year ago
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    Are you talking about when 9-8x=0

  25. anonymous
    • one year ago
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    Because that wont satisfy the problem at a glance

  26. anonymous
    • one year ago
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    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. dinorap1
    • one year ago
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    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
    • one year ago
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    AHHHH yes great point

  29. anonymous
    • one year ago
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    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. dinorap1
    • one year ago
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    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
    • one year ago
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    Yes you can start there

  32. dinorap1
    • one year ago
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    And I did it right? It's correct to get a fraction, right?

  33. anonymous
    • one year ago
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    So we have \[9|9-8x| = 2x+3 \\ \rightarrow \ \ \|9-8x| = \frac{2}{9}x+\frac{1}{3} \]

  34. anonymous
    • one year ago
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    ignore the double bar (typo)

  35. anonymous
    • one year ago
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    So we have \[9|9-8x| = 2x+3 \\ \rightarrow \ \ \ |9-8x| = \frac{2}{9}x+\frac{1}{3} \]

  36. dinorap1
    • one year ago
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    What?? You can divide 9 into 2x too? Man, I never thought you could do that...

  37. dinorap1
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    So just let a be 1/9 and your'e golden :D

  40. anonymous
    • one year ago
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    Ok I will do an example problem and then I want you to try your'e problem:

  41. dinorap1
    • one year ago
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    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
    • one year ago
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    For: \[ |2x-12|\] If \[ 2x-12<0\] Then \[-(2x-12)>0\] ALWAYS Please take a second to consider this

  43. anonymous
    • one year ago
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    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
    • one year ago
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    So: \[|-(2x-12)|=-(2x-12)\]

  45. anonymous
    • one year ago
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    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. dinorap1
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Does this make sense?

  50. dinorap1
    • one year ago
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    Alright... I'm still a little confused, but I'll try and solve it now

  51. anonymous
    • one year ago
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    Bingo... thats the spirit... sometimes it just takes the courage to plow ahead and see what happens :D

  52. anonymous
    • one year ago
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    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. dinorap1
    • one year ago
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    Alright, so when you first start off with |9-8x|=2/9x+1/3 |9-8x| becomes -(9-8x)? On both sides?

  54. anonymous
    • one year ago
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    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
    • one year ago
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    Testing...

  56. dinorap1
    • one year ago
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    Alright, I will. Sooo... only do -(9-8x) on the positive side?

  57. anonymous
    • one year ago
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    OK whew.... If I randomly stop talking its because open study has locked up on me AGAIN today

  58. anonymous
    • one year ago
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    Hopefully this doesnt happen but be warned open study has not cooperated with me today

  59. dinorap1
    • one year ago
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    I'm so sorry for my confusion. Math is definitely not my easiest class...

  60. anonymous
    • one year ago
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    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
    • one year ago
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    Gotta work it then... Just like muscles... If you dont work them they will never grow and stay weak

  62. dinorap1
    • one year ago
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    Alright, I'll do that!

  63. dinorap1
    • one year ago
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    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
    • one year ago
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    For which case?

  65. dinorap1
    • one year ago
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    |9-8x|=2/9x+1/3

  66. dinorap1
    • one year ago
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    Sorry I forgot to type it with parentheses: |9-8x|=(2/9)x+(1/3)

  67. anonymous
    • one year ago
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    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
    • one year ago
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    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. dinorap1
    • one year ago
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    I don't know honestly. What do you mean by that?

  70. dinorap1
    • one year ago
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    I'm ready to give up at this point... this is getting too complicated! :(

  71. anonymous
    • one year ago
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    Sorry I got hung up for a second

  72. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    That symbol btw ==> is just my way of doing an arrow... you read it as "implies"

  75. dinorap1
    • one year ago
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    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
    • one year ago
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    AHHHH Im soooo sorry I made a typoe please hold

  77. anonymous
    • one year ago
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    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. dinorap1
    • one year ago
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    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. dinorap1
    • one year ago
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    *going to

  80. anonymous
    • one year ago
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    It is positive what do you mean?

  81. anonymous
    • one year ago
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    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
    • one year ago
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    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

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