MHB Word problem: distance traveled

AI Thread Summary
The discussion revolves around solving a word problem involving distance traveled by two objects based on their revolutions. The user initially derives equations for distance covered by each object and finds a relationship between their revolutions. They arrive at a solution that matches the book's answer but question why plugging their derived value back into the first equation yields a different result. Another participant clarifies that plugging the value into the first equation actually confirms the consistency of the derived answer. Ultimately, both participants agree that the calculations align correctly, resolving the user's confusion.
NotaMathPerson
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Here' s my attempt

$r=$revolution taken by first
$r+c=$ revolutiin taken by 2nd

$d=a(r+c)$ distance covered by first (1)
$d=br$ distance covered by 2nd (2)

$br=a(r+c)$ solving for r $r=\frac{ac}{b-a}$ and plug into (2)

I get $a=\frac{abc}{b-a}$ this answers agrees with the key answer in the book.

But my question is why do I get different answer when I plug $r$ to eqn. (1)?

Thanks!
 

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NotaMathPerson said:
Here' s my attempt

$r=$revolution taken by first
$r+c=$ revolutiin taken by 2nd

$d=a(r+c)$ distance covered by first (1)
$d=br$ distance covered by 2nd (2)

$br=a(r+c)$ solving for r $r=\frac{ac}{b-a}$ and plug into (2)

I get $a=\frac{abc}{b-a}$ this answers agrees with the key answer in the book.

You mean $d=\frac{abc}{b-a}$, don't you?
NotaMathPerson said:
But my question is why do I get different answer when I plug $r$ to eqn. (1)?

Plugging $r$ into (1) we get the following:
$$d=a(r+c) =a\left (\frac{ac}{b-a}+c\right )=a\left (\frac{ac+c(b-a)}{b-a}\right )=a\frac{ac+cb-ac}{b-a}=a\frac{bc}{b-a}=\frac{abc}{b-a}$$

So, we get the same result.
 
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