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automaticloveletter

  • one year ago

Which of the following is a step that can be used to find the solution to the set of equations? 2d + 1 = 3d + 5 2d = 3d + 5 2d + 1 = 3d 2d + 5 = 3d + 1

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  1. automaticloveletter
    • one year ago
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    Equations: Equation A: c = 2d + 1 Equation B: c = 3d + 5 Is it A?

  2. sdfgsdfgs
    • one year ago
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    A is correct :)

  3. automaticloveletter
    • one year ago
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    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

  4. sdfgsdfgs
    • one year ago
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    wat do u think?

  5. automaticloveletter
    • one year ago
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    No clue xc 1st step?

  6. sdfgsdfgs
    • one year ago
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    read them through step by step plz....its not the 1st...

  7. automaticloveletter
    • one year ago
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    2?

  8. sdfgsdfgs
    • one year ago
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    why? whats wrong there?

  9. automaticloveletter
    • one year ago
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    I'm just guessing.

  10. sdfgsdfgs
    • one year ago
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    :( but u guess right - lucky gurl!! lol

  11. automaticloveletter
    • one year ago
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    So it's the second one?

  12. sdfgsdfgs
    • one year ago
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    Step 1: −3(y) = −3(15 − 2z) [Equation A is multiplied by −3.] Step 2: −3y = 15 − 2z [Equation A in Step 1 is simplified.] See anything wrong there??

  13. automaticloveletter
    • one year ago
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    The simplification?

  14. automaticloveletter
    • one year ago
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    @sdfgsdfgs

  15. sdfgsdfgs
    • one year ago
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    yes - something went wrong there...

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