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lilMissMindset

  • 3 years ago

2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2 need to determine A, B, C

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  1. lilMissMindset
    • 3 years ago
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    @AnimalAin this

  2. AnimalAin
    • 3 years ago
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    We do it the same way. Let me work for a second here...

  3. AnimalAin
    • 3 years ago
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    \[2C + 3B +6Cx+2A+2Bx+2Cx^2=12x^2 \implies\]\[(2A+2C + 3B) +(6C+2B)x+2Cx^2=12x^2 \implies\]\[2C=12 \implies C=6\]\[6C+2B=0 \implies B=-18\]\[2A+2C + 3B=0 \implies A=9\]

  4. AnimalAin
    • 3 years ago
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    The corresponding terms on the left and right sides of the equation have to be the same. So, we match the x^2 coefficients to get C=6, then the x coefficients (note there is none on the RHS, so the LHS has to be zero) to get B=-18, then follow the same pattern to get A=9.

  5. lilMissMindset
    • 3 years ago
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    hey, check A, i think it's 21.

  6. lilMissMindset
    • 3 years ago
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    really, i'm getting it now, thanks a lot.

  7. AnimalAin
    • 3 years ago
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    I think it's nine.... 3*6-2*18=-18 implies that A has to be pretty close to nine.

  8. lilMissMindset
    • 3 years ago
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    you interchanged the value of A and B.

  9. AnimalAin
    • 3 years ago
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    OOPs!! Hate it when that happens!! Bad substitution!!

  10. lilMissMindset
    • 3 years ago
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    i noticed as well. it's alright! :D

  11. AnimalAin
    • 3 years ago
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    You seem to have it now.... even if my arithmetic is lacking. LOL Do math every day.

  12. lilMissMindset
    • 3 years ago
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    you're still one great teacher! :)

  13. lilMissMindset
    • 3 years ago
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    sure, i'll do a lot of math everyday. ammm, do you do some DE's?

  14. AnimalAin
    • 3 years ago
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    A little. What do you have in mind?

  15. lilMissMindset
    • 3 years ago
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    i'll post it on a new one, so that i can give you another medal.

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