May 30th, 2014, 06:24 AM | #11 |
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 | |
May 31st, 2014, 09:13 AM | #12 |
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,299 Thanks: 1023 |
Hmmm...Todd told us this: "I'm a chef for 20 years and now attending school for Hospitality and Restaurant Management. I use to love math in high school. I am just now getting reintroduced to Algebra and would appreciate any help I could get." I don't think exotic solving ways is what he wants/needs; just sufficient to help him pass that HRM course. Using Greg's approach (my favorite!): 9A + 10B + 12C = 398 [1] 6A + 4B + 4C = 164 [2] A + 2B + C = 58 [3] Using elimination of variables, it can readily be determined that A = 6, B = 20 and C = 12. I'm pretty sure Todd wants to "remember" how above is accomplished. The easiest variable to eliminate first is C (2 multiplications only); multiply [2] by 3 and [3] by 12, and we now have: 09A + 10B + 12C = 398 [1] 18A + 12B + 12C = 492 [2] 12A + 24B + 12C = 696 [3] Subtract [1] from [2]: 9A + 2B = 94 [4] Subtract [1] from [3]: 3A + 14B = 298 [5] Multiply [5] by 3: 9A + 42B = 894 [5] Subtract [4] from [5]: 40B = 800 : so B = 800/40 = 20 Now you can go back and substitute B=20; let's take [4]: 9A + 2(20) = 94 9A + 40 = 94 9A = 54 A = 6 Now pick one of [1],[2],[3] and substitute A=6 and B=20 to get C : all yours to try! Last edited by Denis; May 31st, 2014 at 09:17 AM. |
June 4th, 2014, 03:07 AM | #13 | |
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions | Quote:
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