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anonymous

  • 5 years ago

Find the answer. B needs to be by itself. (a+b)/2c = b+d

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  1. anonymous
    • 5 years ago
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    You want to solve for \(b\)?

  2. anonymous
    • 5 years ago
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    yes!

  3. anonymous
    • 5 years ago
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    And show your work, please!

  4. anonymous
    • 5 years ago
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    OK. Start by multiplying both sides by \(2c\) to get rid of the denominator. So, we will have \(a+b=(b+d)2c\). Opening the parenthesis gives \(a+b=2bc+2cd\). Since we're trying to solve for \(b\), we should combine all terms with \(b\) in one side and everything else on the other side. That gives \(b-2bc=2cd-a\). Take \(b\) as a common factor in the left hand side, you get \(b(1-2c)=2cd-a\). Finally divide both sides by \(1-2c\) you get \(b=\frac{2cd-a}{1-2c}\).

  5. anonymous
    • 5 years ago
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    b = (a-2*c*d)/(2*c-1)

  6. anonymous
    • 5 years ago
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    cross multiply then a+b =2c(b+d) then take b containing term on one side and rest on other side so b(2c-1) = a-2cd now divide both side by (2c-1) so u get b= (a-2cd)/(2c-1)

  7. anonymous
    • 5 years ago
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    (a+b)/2c = b=d 2bc + 2cd = a + b .................. (multiply both side by 2c) 2bc - b = a - 2cd ...................... (changing side of b) b ( 2c - 1) = a - 2cd b = (a-2cd) / (2c - 1)

  8. anonymous
    • 5 years ago
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    Everyone had good answers! Thanks, and I agree with them.

  9. anonymous
    • 5 years ago
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    Well they all are actually the same :)

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