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sammy5592 Group Title

Part 1: Solve the following system of equations. Please show your work to receive full credit. x - y = 10 2x + y = 2 Part 2: Explain which method you chose to solve the system and why you felt it was the best choice.

  • one year ago
  • one year ago

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  1. hartnn Group Title
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    Add those two equations first and tell me what u get?

    • one year ago
  2. Mellissa Group Title
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    combine like terms first

    • one year ago
  3. sammy5592 Group Title
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    3x+y=12

    • one year ago
  4. Mellissa Group Title
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    is that Part 2 part of the equation ?@Sammy5592

    • one year ago
  5. sammy5592 Group Title
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    no @Mellissa

    • one year ago
  6. shaik0124 Group Title
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    |dw:1346076054334:dw|

    • one year ago
  7. Mellissa Group Title
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    ok @Sammy

    • one year ago
  8. shaik0124 Group Title
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    -3y=18 y=-6 x-(-6)=10 x+6=10 x=10-6 x=4

    • one year ago
  9. shaik0124 Group Title
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    therefore x=4,y=-6 is answer

    • one year ago
  10. sammy5592 Group Title
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    can u write the firsrt 1 out like that

    • one year ago
  11. hartnn Group Title
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    @sammy5592 how did u get 3x+y=12 u should get 3x=12 check again.

    • one year ago
  12. shaik0124 Group Title
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    i multiplied first equation by 2 then subtracted from 2nd equation ..this gives y .put y value in any one of the equations u will get x,theefore x,y u got is solution

    • one year ago
  13. sammy5592 Group Title
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    no i mean how you did the second equation

    • one year ago
  14. shaik0124 Group Title
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    did u got @sammy5592

    • one year ago
  15. sammy5592 Group Title
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    ???

    • one year ago
  16. shaik0124 Group Title
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    did u understand the solution i replied for above problem

    • one year ago
  17. sammy5592 Group Title
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    yes

    • one year ago
  18. sammy5592 Group Title
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    part 2 ???

    • one year ago
  19. shaik0124 Group Title
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    it is substitution method

    • one year ago
  20. shaik0124 Group Title
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    it is best bcoz if u find 1 value then automatically it is easy to find another value

    • one year ago
  21. sammy5592 Group Title
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    Thanks :)

    • one year ago
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