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November 10th, 2010, 05:47 PM   #1
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Can this even be solved without more info?

Hi, haven't done maths since highschool, any help would be appreciated.
Is it possible to solve for A & B if I know C? (& if it helps that all numbers are natural, like 1 or 2 or 3 or 4 not like 2.something)

6A+B=C (Example: 6A+B=52)

We also know -

A=DE
B=D+E-2

PS. Don't know if this is posted in the right section.
Thanks.
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November 10th, 2010, 05:53 PM   #2
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Re: Can this even be solved without more info?

It appears you have five unknowns, but only three equations, so you cannot find a numerical value for the unknowns. The best you can do is express the unknowns in terms of the others.

And yes, this is the appropriate sub-forum for this question.
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November 10th, 2010, 07:08 PM   #3
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Re: Can this even be solved without more info?

Thanks for the quick reply.

Question: I have four unknowns not 5 (C is known) but does that still mean I need 4 equations if I want to solve it? Thanks again.
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November 10th, 2010, 07:21 PM   #4
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Re: Can this even be solved without more info?

Sorry, didn't catch that C was known.

But yes, you need the same number of independent equations as you have unknowns to completely solve the system.
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November 11th, 2010, 04:43 AM   #5
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Not necessarily.

One has 6DE + D + E - 2 = C, so D = (C + 2 - E)/(6E + 1) = C + 2 - (6C + 13)E/(6E + 1).
If, say, C = 52, D = 54 - (13)(25)E/(6E + 1).
Since D and E are natural numbers, one can now see that E must be 2 or 4.
Hence D and E are 2 and 4 (in either order) and so A = 8 and B = 4, and this particular case has a unique solution.
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November 11th, 2010, 11:14 AM   #6
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Re: Can this even be solved without more info?

Wow thanks so much SkipJack! That is amazing!

I'm not very good at Algebra (understatement) though so I got a bit lost inbetween some of your steps, would you or someone else be able to fill in these gaps for me please..

1. You say D = C + 2 - (6C + 13)/(6E + 1), then when you substitute C=52 into the equation it becomes D = 54 - (13)(25)E/(6E + 1).

How did you get the (13)(25)E, When I substitute C I get D=54-(325)/(6E+1) ?


2. Then from D = 54 - (13)(25)E/(6E + 1) you say 'since D & E are natural numbers one can now see that E must be 2 or 4.'

How do you see that E must be 2 or 4, or how do you work out?

Thanks again guys!
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November 11th, 2010, 07:24 PM   #7
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I had omitted an "E" in the numerator, which I've now corrected. However, the "E" was not omitted when I did the substitution of C = 52, so D = 54 - (13)(25)E/(6E + 1) was correct.
I obtained (13)(25) by first calculating 6C + 13 = 6(52) + 13 = 325 = (13)(25).

Since E and 6E + 1 cannot have a common divisor greater than 1, (13)(25)E/(6E + 1) can be a natural number only if 6E + 1 is a divisor of (13)(25).
That implies that 6E + 1 = 13 or 25 or (13)(25), so E = 2 or 4 or 54.
One needn't consider 6E + 1 = 5 or 13(5), as E is then not a natural number.
Also, E = 54 is ruled out as it implies D = 0.
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November 17th, 2010, 08:28 AM   #8
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Re: Can this even be solved without more info?

People don't try you have 4 unknown and 3 equations as Mark said it's impossible to solve if there were another 4 equations you could.
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November 19th, 2010, 12:18 PM   #9
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That's incorrect. If the solutions are required to be integers, fewer equations (than variables) may well suffice.
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