I forgot my old school math. Someone throw me a bone on this one. Would prefer to see it worked out. A+B=65 A3+B2=179

A + B = 65 3*A + 2*B = 179 A = 65-B substitute (65-B)*3 + 2*B = 179 195 - 3*B + 2*B = 179 195 - B = 179 B = 16 substitute again A = 65-16 A = 49 Easy

Could be done via matrices with cramers rule as well imho is a lot easier than the substitution method.

ahhhh i had the right path it was always the end that screwed me my second equation i solved A = 49 but then when i ran it back through i divided by 2 twice and ended up with 147+B2=179 B=8 the first and third time i ended up with 195-B3+B2=179 195-B5=179 -B5 = 16 B = 3.2 I thought I was making my beginning formulas wrong. Original question... don't ask... A farmer has three legged cows and chicken. There are 65 heads and 179 legs. How many cows does the farmer have. Is this the other method you were refering to CRXYEM

I suppose- it's been too long though... plus I would normally just do it in my head via brute force in about 5-6 seconds- but that wouldn't give anything to go by in terms of "showing the work" in the thread. I really don't think it's anywhere near complex enough to use any specific method to solve it.