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November 10th, 2010, 04:47 PM  #1 
Newbie Joined: Nov 2010 Posts: 3 Thanks: 0  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+E2 PS. Don't know if this is posted in the right section. Thanks. 
November 10th, 2010, 04:53 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs  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 subforum for this question. 
November 10th, 2010, 06:08 PM  #3 
Newbie Joined: Nov 2010 Posts: 3 Thanks: 0  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. 
November 10th, 2010, 06:21 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs  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. 
November 11th, 2010, 03:43 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 18,716 Thanks: 1532 
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. 
November 11th, 2010, 10:14 AM  #6 
Newbie Joined: Nov 2010 Posts: 3 Thanks: 0  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! 
November 11th, 2010, 06:24 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,716 Thanks: 1532 
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. 
November 17th, 2010, 07:28 AM  #8 
Newbie Joined: Nov 2010 Posts: 9 Thanks: 0  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.

November 19th, 2010, 11:18 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 18,716 Thanks: 1532 
That's incorrect. If the solutions are required to be integers, fewer equations (than variables) may well suffice.


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