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anonymous

  • 5 years ago

|7-3b|= |5b+15|

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  1. anonymous
    • 5 years ago
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    Use the definition of absolute value; that is, the absolute value of a quantity, y, is defined by,\[|y|=\sqrt{y^2}\]So, for your problem, we have,\[\sqrt{(7-3b)^2}=\sqrt{(5b-15)^2}\]You can then square both sides to get\[(7-3b)^2=(5b-15)^2\]Expand and collect your terms into a quadratic equation\[16b^2-108b+176=0\]and solve for b:\[b=4, \frac{11}{4}\]

  2. anonymous
    • 5 years ago
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    i am trying to find the possible values for the inequality in this problem.....i didnt think it was a quadratic equation.

  3. anonymous
    • 5 years ago
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    You posted an equality...

  4. anonymous
    • 5 years ago
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    The solutions are b = -1 and b = -11. Lokisan put 5b - 15 instead 5b + 15 which led to incorrect answers. I did it by breaking the absolute value into separate equations. I'll do it this way: |7 - 3b| = |5b + 15| --> 7 - 3b = |5b + 15| OR 7 - 3b = -|5b + 15|. Equation A: 7 - 3b = |5b + 15| 5b + 15 = 7 - 3b --> 8b = -8 --> b = -1 5b + 15 = -(7 - 3b) = 3b - 7 --> 2b = -22 --> b = -11 Equation B: 7 - 3b = -|5b + 15| OR 3b - 7 = |5b + 15| 5b + 15 = 3b - 7 --> 2b = -22 --> b = -11 5b + 15 = 7 - 3b --> 8b = -8 --> b = -1

  5. anonymous
    • 5 years ago
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    Yeah, my bad.

  6. anonymous
    • 5 years ago
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    oh wow....thats a long problem lol but i will try it, thank you both

  7. anonymous
    • 5 years ago
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    and i was also thinking I had to set them both to zero because I had an example that looked the same that did that

  8. anonymous
    • 5 years ago
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    It's best to use either the definition of absolute value (like I did) or branlegr's case method.

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