- #1

Ravian

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(AB+1)/(ABC+A+C)=0.138, Find values of A,B,C.

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- Thread starter Ravian
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- #1

Ravian

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(AB+1)/(ABC+A+C)=0.138, Find values of A,B,C.

- #2

symbolipoint

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Does that example have more information with it? You show one equation built with three unknown variables. Two more equations are necessary as requirement for finding VALUES for A, B, and C.

- #3

Ravian

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- #4

JonF

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If you need ALL values you just need to treat one variable as a dependant variable, so solving for A in terms of B and C (if I did my algebra right) you get:

A = (0.138C - 1)/(B -0.138ABC - 0.138). So for any B,C provided ABC+A+C ≠ 0

That's as spefic as you can get i believe.

- #5

JonF

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Isolate all terms with an “A” to one side, then factor out the “A”. would be a good way to aproach this.

- #6

symbolipoint

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Nonsense. You only have one equation which uses three variables. You have no other information. You cannot find any values for the variables. The best you can do is solve for each variable to find a formula for each. Otherwise, you cannot solve for any values.

- #7

mathplease

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(AB+1)/(ABC+A+C)=0.138, Find values of A,B,C.

without any context behind the question it's reasonable to interpret it in another popular way. although i still don't think it works with this particular question you posted here, but consider the similar type:

(AB + 1) / (CBA + A + B) = 0.138

now instead of treating AB as A multiplied with B, for example, we can interpret it as the integer number with A tens and B units. in this case it has a unique solution:

A = 6, B = 8, C = 4

(68 + 1) / (486 + 6 + 8) = 0.138

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