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ashleyb

Solve the following system of equations. 2a – 3b + c = 10 2a – 2b – 2c = 2 a + 3b + 2c = –1

  • one year ago
  • one year ago

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  1. ghazi
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    use determinant to solve above equation

    • one year ago
  2. ashleyb
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    idk wat that means

    • one year ago
  3. amistre64
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    there are different methods that all come to the same results; since you have not defined a method, the determinant is suitable

    • one year ago
  4. daren13354
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    the determinant is kidna useless you can tell if it's not factorable once you have the equation narrowed to 2 variables.

    • one year ago
  5. daren13354
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    just use substitution :D!

    • one year ago
  6. amistre64
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    ugh, substitution on these takes forever!!

    • one year ago
  7. amistre64
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    id eliminate :)

    • one year ago
  8. ashleyb
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    how do u eliminatte

    • one year ago
  9. amistre64
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    you scale the eqauations so that you can combine them and eliminate one of more of the variables

    • one year ago
  10. amistre64
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    R1, 2a – 3b + c = 10 R2, 2a – 2b – 2c = 2 R3, a + 3b + 2c = –1 R3(-2), R2(-1) R1, 2a – 3b + c = 10 R2, -2a + 2b + 2c =-2 R3, -2a -6b -4c = 2 R1+R2, and R1+R3 R1, 2a – 3b + c = 10 R2, 0 - b + 3c = 8 R3, 0 - 9b - 3c = 12 R2(-9) R1, 2a – 3b + c = 10 R2, 0 +9b -27c = -72 R3, 0 - 9b - 3c = 12 R2 + R3 R1, 2a – 3b + c = 10 R2, 0 +9b -27c = -72 R3, 0 + 0 - 30c =-60

    • one year ago
  11. amistre64
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    -30c = -60 c = 2

    • one year ago
  12. amistre64
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    R1, 2a – 3b + (2) = 10 R2, 0 +9b -27(2) = -72 R1, 2a – 3b = 10-2 R2, 0 +9b = -72+54 b = -18/9 = -2 R1, 2a – 3(-2) = 8 R1, 2a + 6 = 8 a = (8-6)/2 = 1

    • one year ago
  13. amistre64
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    2a – 3b + c = 10 1 -2 2 ------------- 2 + 6 + 2 = 10 2a – 2b – 2c = 2 1 -2 2 -------------- 2 + 4 - 4 = 2 a + 3b + 2c = –1 1 -2 2 ------------------ 1 -6 + 4 = -1

    • one year ago
  14. satellite73
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    wow that is a lot of work i do not like these so much do you?

    • one year ago
  15. amistre64
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    i find them tedious

    • one year ago
  16. amistre64
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    it takes longer to show the work than it does to actually get the answer :)

    • one year ago
  17. ashleyb
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    thankyou for ur help!

    • one year ago
  18. amistre64
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    i hope it was helpful :)

    • one year ago
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