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angela210793

  • 5 years ago

Simplify: |1-a+b|-|a-b| -------------- |b-a| +a

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  1. anonymous
    • 5 years ago
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    \[\frac{1-a+b-a+b}{b-a+a} = \frac{1-2a + 2b}{b}\] all you've got to do is simplify by multiplying the sign to the values in brackets and compute ^_^

  2. angela210793
    • 5 years ago
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    those aren't brackets :):):):) absolute value :S

  3. anonymous
    • 5 years ago
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    hmm , oh right, you're right lol. If you have absolute values, then the signs switch.

  4. anonymous
    • 5 years ago
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    not sure how though? >_< I'm sorry. I know that |b-a| = a-b

  5. anonymous
    • 5 years ago
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    but not sure of the one in the denominator.

  6. angela210793
    • 5 years ago
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    i'm not sure how to solve this 2.....thnx a lot anyway :) ^_^

  7. anonymous
    • 5 years ago
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    numerator*

  8. anonymous
    • 5 years ago
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    Alright, let me see if I can make it simpler, |4x| = 4x, |-4x| = 4x |4x-3|= 3-4x |2a +b| = 2a+b makes any sense?

  9. angela210793
    • 5 years ago
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    yea it does...thnx :)so i just switch signs...nothing else?

  10. anonymous
    • 5 years ago
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    looks like it :) np dear ^_^

  11. anonymous
    • 5 years ago
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    you switch signs only when you have a negative inside with another variable, others stay the same ^_^

  12. angela210793
    • 5 years ago
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    hmm how abt -|a-b|???? is it a-b?

  13. anonymous
    • 5 years ago
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    first find |a-b| = b-a then multiply it by (-) = -(b-a) = a-b LOL it's the same thing :)

  14. anonymous
    • 5 years ago
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    so yes ^_^

  15. angela210793
    • 5 years ago
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    ^_^ Tjnx a lot :):):) ^_^

  16. anonymous
    • 5 years ago
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    np dear :)

  17. angela210793
    • 5 years ago
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    thnx*

  18. anonymous
    • 5 years ago
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    For equations having terms with absolute value, you have to calculate twice, once if the value inside the brackets is positive (non-negative in a broader sense), and then if the value inside the brackets is negative. So, you have to think about cases, when (a-b)<0, when 0<=(a-b)<=1 and when (a-b)>1. Now the solution. 1st case: when (a-b)<0 or a<b, |a-b| = b-a. And, |1-a+b| = |1-(a-b)| = 1-a+b. |b-a| = b-a. So, the simplification will be 1/b. 2nd case: when 0<=(a-b)<=1, then |a-b| = a-b, |1-a+b| = |1-(a-b)| = 1-a+b. |b-a| = a-b. So the simplification will be (1-2a+2b)/(2a-b). 3rd case: when (a-b)>1, then |a-b| = a-b, |1-a+b| = |1-(a-b)| = a-b-1, |b-a| = a-b. So the simplification will be 1/(b-2a). Enjoy. :)

  19. angela210793
    • 5 years ago
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    Oh thanks a lot :):):)

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