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anonymous

  • one year ago

1/7(B+35)3 linear equations

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  1. anonymous
    • one year ago
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    @Leong

  2. anonymous
    • one year ago
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    @ganeshie8

  3. anonymous
    • one year ago
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    @LynFran

  4. anonymous
    • one year ago
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    @Leong

  5. anonymous
    • one year ago
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    @mathway

  6. anonymous
    • one year ago
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    @izuru

  7. anonymous
    • one year ago
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    @mathstudent55

  8. anonymous
    • one year ago
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    can you help me @RunawayGalaxy

  9. anonymous
    • one year ago
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    Well you're not being too clear in the question you're asking here, do you want to solve for B ?

  10. anonymous
    • one year ago
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    yes

  11. mathstudent55
    • one year ago
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    I don't understand what the problem is. There is no equal sign, so this is not an equation. what do you need to do?

  12. anonymous
    • one year ago
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    ohh so sorry =3

  13. mathstudent55
    • one year ago
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    Ok. Let me write it again.

  14. anonymous
    • one year ago
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    ok

  15. mathstudent55
    • one year ago
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    Solve for B: \(\large \dfrac{1}{7}(B + 35) = 3\)

  16. mathstudent55
    • one year ago
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    Is that the problem?

  17. anonymous
    • one year ago
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    yes

  18. mathstudent55
    • one year ago
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    Ok. You have a multiplication of 1/7 on the left side that we need to get rid of. To get rid of a multiplication, you multiply by its reciprocal. The reciprocal of 1/7 is 7, so we multiply both sides by 7. Can you do that and show what you get?

  19. mathstudent55
    • one year ago
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    \(\large 7 \times \dfrac{1}{7}(B + 35) = 7 \times 3\) What do you get?

  20. anonymous
    • one year ago
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    7b and 245

  21. mathstudent55
    • one year ago
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    No. On the let side, what is 7 * 1/7? The whole idea of multiplying both sides by 7 is to get rid of 1/7 on the left side. What is 7 * 1/7?

  22. anonymous
    • one year ago
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    1

  23. mathstudent55
    • one year ago
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    |dw:1439771895851:dw|

  24. mathstudent55
    • one year ago
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    Correct. That means on the left side, we only have now: \(\large B + 35\) On the right side, what is 7 * 3?

  25. anonymous
    • one year ago
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    21

  26. mathstudent55
    • one year ago
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    Correct. That means now we have \(\large B + 35 = 21\)

  27. mathstudent55
    • one year ago
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    We want B alone, but we have B + 35 on the left side. Since plus 35 means add 35, we do the opposite to get rid of the +35. We subtract 35. We must do the same to both sides of an equation, so we subtract 35 from both sides of the equation. \(\large B + 35 - 35 = 21 - 35\) What do you get?

  28. anonymous
    • one year ago
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    35-21=14 B=14

  29. anonymous
    • one year ago
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    thanks. :)

  30. anonymous
    • one year ago
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    -14

  31. mathstudent55
    • one year ago
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    Be careful. The left side is correct. B + 35 - 35 is simply B. On the right side, you have 21 - 35, not 35 - 21. The answer is not B = 14

  32. anonymous
    • one year ago
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    lol

  33. mathstudent55
    • one year ago
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    Oh, I see. You got it. The answer is B = -14

  34. anonymous
    • one year ago
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    Sorry I didn't help, stepped away from the computer

  35. anonymous
    • one year ago
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    its fine took an hour to get help :P

  36. mathstudent55
    • one year ago
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    Now we check the answer in the original equation to see if it is correct. We get B = -14 If we plugin -14 for B in the original equation, we must get a true statement. \(\dfrac{1}{7}(B + 35) = 3\) \(\dfrac{1}{7}(-14 + 35) = 3\) \(\dfrac{1}{7}(21) = 3\) \(3 = 3\) Since 3 = 3 is a true statement, our solution B = -14 is correct.

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