MHB Word problem: distance traveled

[NotaMathPerson]

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!

 

[mathmari]

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?

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.