I need help!

Osiris

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Feb 17, 2019
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The 3 numbers attached to each animal needs to be between 1-9.

In the attachment is the problem I have been posed.

I need the values of each animal

I cannot figure it out for the life of me
 

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mmm4444bot

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The [number for] each animal needs to be between 1-9 …
Hello. You also know that the numbers must be Whole numbers (from 1 through 9). In your attempts, did you try algebra? Did you guess and check?

There isn't enough information given, to find all three numbers (directly) from solving equations. However, we could use algebra to find formulas for two of the unknowns -- each in terms of the remaining unknown. After that, we would use trial and error, to reason out the solution. Start by picking symbols for the unknown numbers.

Let B = the number of bats
Let C = the number of chickens
Let R = the number of rabbits

Choosing to find formulas in terms of B, for example, might yield something like this (made-up information):

C = 2B

R = B/2

If this were true (and it isn't), then we could see right away that B must be an even number.

From there, we would start guessing and checking -- trying B=2, B=4, B=6, etc. -- until we find Whole numbers for both C and R.

Please show your work, if you've tried anything. Check out the forum's submission guidelines, too. Cheers 🐬
 
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HallsofIvy

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Jan 27, 2012
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Using B, C, and R for the number of bats, chickens, and rabbits, respectively,
"five times the number of chickens I have plus three times the number of bats equals six times the number of rabbits": 5C+ 3B= 6R. We could also write that as 5C+ 3B- 6R= 0.
"The number of chickens plus twelve times the number of rabbits is equal to six times the number of bats I have": C+ 12R= 6B. We could also write that as C- 6B+ 12R= 0.

Seeing "3B" in one equation and "-6B" in the other I would multiply the first equation by 2 and add them: 10C+ 6B- 12R+ C- 6B+ 12R= 11C= 0 so C= 0. That contradicts the requirement that all the number be between 1 and 9. If that really is a requirement then there is no solution! I am going to assume that the solutions must be from 0 to 9.

With C= 0, the two equations reduce to 3B- 6R= 0 and -6B+ 12R= 0. Those both reduce to B= 2R.

With only two equations in 3 unknowns, we cannot find a single solution.
We can however, calculate all the solutions with all numbers from 0 to 9 by looking at possible values for R.
If R= 0 then B= 2(0)= 0 so one solution is that there are no animals at all!
If R= 1 then B= 2(1)= 2 so another solution is no chickens, 1 rabbit, and 2 bats.
If R= 2 then B= 2(2)= 4 so another solution is no chickens, 2 rabbits, and 4 bats.
If R= 3 then B= 2(3)= 6 so another solution is no chickens, 3 rabbits, and 6 bats.
If R= 4 then B= 2(4)= 8 so another solution is no chickens, 4 rabbits, and 8 bats.

If R is 5 or larger then B is larger than 9 so those are the only possible solutions.
 
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