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

• Oct 18th 2010, 09:39 AM
JRichardson1729
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:

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

I believe the solution to be $\displaystyle A$ apples$\displaystyle =\frac{AbC}{aB}$ cherries. The problem is that I am not sure whether it is acceptable to have $\displaystyle A$ 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.
• Oct 18th 2010, 09:53 AM
Wilmer
Quote:

Originally Posted by JRichardson1729
I believe the solution to be $\displaystyle A$ apples$\displaystyle =\frac{AbC}{aB}$ 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...
• Oct 18th 2010, 11:27 AM
JRichardson1729
Quote:

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 $\displaystyle A$ 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 $\displaystyle x, y, z$ respectively.

$\displaystyle ax=by, By=Cz$

$\displaystyle aBx=bBy, bBy=bCz$

$\displaystyle aBx=bCz$

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

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

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 $\displaystyle A$ on one side (with $\displaystyle x$ i.e. the weight of an apple).
• Oct 21st 2010, 08:12 AM
JRichardson1729
Bump.
• Oct 21st 2010, 08:41 AM
Unknown008
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.