Math Help

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reckedracing

TTIWWOP
VIP
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
There are horses and three legged cows there are a total of 63 heads and 240 legs how many horses?

The easiest way is to substract three times the number of heads from the legs and that will give you the amount of horses. The reasoning is as follows:

legs = l
heads = h
horses = x
cows = (h - x)
l = 3(h-x)+4x
l = 3h-3x+4x
l-3h = x
x = l-3h


"On a farm there are horses and three-legged-cows. There are total of 100 heads and 376 legs. How many horses are on the farm?"

100 * 4= 400
400 - 376 =24
100 - 24 = 76
76 horses 24 cows 24 * 3 = 72 (cow legs) 76 * 4 = 304 (horses legs)
 
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Could be done via matrices with cramers rule as well imho is a lot easier than the substitution method.

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.
 
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