# Thread: An acceptable answer for a question that relates to simultaneous equations?

1. ## An acceptable answer for a question that relates to simultaneous equations?

I recently happened upon a problem. I have produced a solution, however, I am not sure whether it is acceptable or not. Here is the problem:

$a$ apples weigh the same as $b$ bananas. $B$ bananas weigh the same as $C$ cherries. How many cherries weigh the same as $A$ apples?

I believe the solution to be $A$ apples $=\frac{AbC}{aB}$ cherries. The problem is that I am not sure whether it is acceptable to have $A$ on both sides of the equation. I have asked a few individuals and opinions vary. Is the solution acceptable?

Thank you. All help on this problem is appreciated.

2. Originally Posted by JRichardson1729
I believe the solution to be $A$ apples $=\frac{AbC}{aB}$ cherries.
If A = AbC / aB
then aB = AbC / A
so aB = bC

I'll leave it at that, since the problem makes no sense to me...

3. Originally Posted by Wilmer
If A = AbC / aB
then aB = AbC / A
so aB = bC

I'll leave it at that, since the problem makes no sense to me...
Unfortunately, the question asks how many cherries weigh the same as $A$ apples, so I think that it is necessry to have an A somewhere in the solution.

This is my method:

Posit that the weights of an apple, banana and cherry are $x, y, z$ respectively.

$ax=by, By=Cz$

$aBx=bBy, bBy=bCz$

$aBx=bCz$

$x=\frac{bC}{aB}z$

$Ax=\frac{AbC}{aB}z$

I'm relatively sure that you need the last step, or at least something like it, but I am not sure about how to only get $A$ on one side (with $x$ i.e. the weight of an apple).

4. Bump.

5. I think it's good.

If you remove the A on both sides, you'll get the number of cheries that will weigh the same as one apple. So, your solution is good. At least, that's what I think.