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12 October, 02:37

A = 2BD + 2BC + 2DC for C

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  1. 12 October, 02:55
    0
    Solve it for C?

    Ok, so A = 2BD + 2BC + 2DC

    First, by trying to isolate the C alone on one side, we get rid of any common terms that aren't C. All terms on the right have a 2, so we factor that 2 out: A = 2BD + 2BC + 2DC

    --> A = 2 (BD + BC + DC)

    Next divide both sides by that 2:

    A/2 = 2 (BD+BC+DC) / 2 - - > A/2 = BD + BC + DC

    Now, we can subtract the BD from the right since it contains no C. Remember again that in "solving for" a variable, that really translates into "isolate it on one side":

    (A/2) - BD = BD - BD + BC + DC

    --> (A/2) - BD = BC + DC

    We're almost there! So on the right we can finally factor out our C, in which [BC+DC] becomes [C (B+D) ]. Let's put it all together:

    (A/2) - BD = BC + DC - - > (A/2) - BD = C (B+D)

    Now we can divide out the (B+D) to get our C alone: [ (A/2) - BD]: (B+D) = C (B+D) : (B+D)

    --> [ (A/2) - BD] / (B+D) = C
  2. 12 October, 02:58
    0
    Start by subtracting 2BD from both sides:

    A-2BD=2BC+2DC

    Factor the C out of the equation:

    A-2BD=C (2B+2D)

    Divide both sides of the equation by 2B+2D:

    (A-2BD) / (2B+2D) = C
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