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Spring98

  • one year ago

The work of a student to solve a set of equations is shown: Equation A: y = 15 – 2z Equation B: 3y = 3 – 4z Step 1: –3(y) = –3(15 – 2z) [Equation A is multiplied by –3.] 3y = 3 – 4z [Equation B] Step 2: –3y = 15 – 2z [Equation A in Step 1 is simplified.] 3y = 3 – 4z [Equation B] Step 3: 0 = 18 – 6z [Equations in Step 2 are added.] Step 4: 6z = 18 Step 5: z = 3 In which step did the student first make an error? Step 4 Step 3 Step 2 Step 1

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  1. Spring98
    • one year ago
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    @satellite73

  2. pooja195
    • one year ago
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    Try solving it yourself then see where he made the mistake.

  3. pooja195
    • one year ago
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    HINT: start with substitution

  4. Spring98
    • one year ago
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    i did! is step one wrong?

  5. pooja195
    • one year ago
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    Yep xD

  6. pooja195
    • one year ago
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    What did he do wrong?

  7. Spring98
    • one year ago
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    so step one is the right answer?

  8. pooja195
    • one year ago
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    Well yeah but what was his mistake?

  9. Spring98
    • one year ago
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    he put a -3 instead of a 3

  10. pooja195
    • one year ago
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    yea but one thing i dont understand why is there another y?

  11. Hero
    • one year ago
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    Are you sure Step 1 is wrong?

  12. pooja195
    • one year ago
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    \[\huge~–3(y) = –3(15 – 2z)\] it says y equals... 15-2z...

  13. Hero
    • one year ago
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    Yes but we're allowed to multiply both sides of an equation by -3

  14. pooja195
    • one year ago
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    It says y=15-2z the person never plugged it in place of y .

  15. Hero
    • one year ago
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    The problem says the student is working to solve the system of equations. There's more than one way to solve a system.

  16. Hero
    • one year ago
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    Multiplying both sides of an equation by -3 does not count as an "error". Nothing "wrong" was done in that step. The student approached differently than you would have.

  17. pooja195
    • one year ago
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    Right but it looks like they used substituion.... :/ This is typed out weirdly .-.

  18. Spring98
    • one year ago
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    so witch one would it be?

  19. Spring98
    • one year ago
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    i'm confused now!

  20. Spring98
    • one year ago
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    plz explain this to me!

  21. Spring98
    • one year ago
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    anyone?

  22. Nnesha
    • one year ago
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    y = 15 – 2z 3y = 3 – 4z how would u find the solution ? you need same coefficient but opposite sign to isolate for one variable that's why multiply 1st equation by -3 so -3(y = 15 -2z) = ???

  23. Spring98
    • one year ago
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    @Hero is the answer step 4

  24. Hero
    • one year ago
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    Guessing is not the best approach to this. How'd you make it through the other steps successfully? The truth is you wouldn't even really need to ask which step is correct, if you had the correct approach. But I understand you're asking for help. So we know step 1 is correct. What are your thoughts on step 2?

  25. Spring98
    • one year ago
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    i don't see anything wrong with step 2

  26. Hero
    • one year ago
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    So the equations in step 1 are equivalent to the equations in step 2? Because when solving systems, for each step, the equations have to all be equivalent.

  27. Spring98
    • one year ago
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    im not sure

  28. Hero
    • one year ago
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    In step 1, for the first equation, after multiplying both sides by -3, we have: –3(y) = –3(15 – 2z) In step 2, that equation was supposedly simplified to get: –3(y) = 15 – 2z Tell me, is –3(y) = 15 – 2z equivalent to y = 15 – 2z ?

  29. Spring98
    • one year ago
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    no

  30. Hero
    • one year ago
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    Why not?

  31. Spring98
    • one year ago
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    becuase its -3y and the other one is plain y

  32. Hero
    • one year ago
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    So what does this observation tell you?

  33. Spring98
    • one year ago
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    that step 2 is wrong. so step two is the answer

  34. Hero
    • one year ago
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    Bingo.

  35. Spring98
    • one year ago
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    thanks for explaning.

  36. Hero
    • one year ago
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    You're most welcome.

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