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Math Help - If C=A+A*B what's A in terms of B & C?

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    If C=A+A*B what's A in terms of B & C?

    I find myself in the odd situation of actually having to use algebra in real life many many years after leaving school!

    If C = A + A * B, how do I find A from knowing B and C? If it makes any difference, A,B & C are all greater than zero.
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by rbbot View Post
    I find myself in the odd situation of actually having to use algebra in real life many many years after leaving school!

    If C = A + A * B, how do I find A from knowing B and C? If it makes any difference, A,B & C are all greater than zero.
    I'll take A*B as A times B.

    A+AB = C, the left hand side of this equation has common factor A, we can therefore write it as:

    A(1+B) = C, then dividing both sides by 1+B,

    A = \frac{C}{1+B}

    Hope this helps
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  3. #3
    Super Member Bacterius's Avatar
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    The thing to remember here, is that your equation a + a \times b = c is equivalent to a \times 1 + a \times b = c, and then you clearly see the common factor. As you practice more and more, it will become a reflex.
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  5. #5
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    Why can't you just subtract the product of (A*B) from C? Would this be wrong?
    I don't think my method would answer his question because it doesn't simply the equation to A in terms of B and C (it is A in terms of A,B, and C) amd I right?

    So...
    C = A + (AB) -> (-AB)
    C - AB = A
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Masterthief1324 View Post
    Why can't you just subtract the product of (A*B) from C? Would this be wrong?
    I don't think my method would answer his question because it doesn't simply the equation to A in terms of B and C (it is A in terms of A,B, and C) amd I right?

    So...
    C = A + (AB) -> (-AB)
    C - AB = A
    You can do that, but where does it get you? The thing here is that we were asked to "find A given B and C". The word "find" implies that we should solve for A.
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  7. #7
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    Quote Originally Posted by Masterthief1324 View Post
    I don't think my method would answer his question because it doesn't simply the equation to A in terms of B and C (it is A in terms of A,B, and C) amd I right?
    correct.
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