Algebra Pre-Algebra and Basic Algebra Math Forum

 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+E-2 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: 466 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 sub-forum 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: 466 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: 18,954 Thanks: 1601 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: 18,954 Thanks: 1601 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: 18,954 Thanks: 1601 That's incorrect. If the solutions are required to be integers, fewer equations (than variables) may well suffice.

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