Question 1194505
<pre>
              SPEED           TIME          DIST.

UPSTR         20-c             t              60

DOWNSTR       20+c             t             100
</pre>
Value of {{{t}}} is equal both to {{{60/(20-c)}}} and to {{{100/(20+c)}}}, so meaning {{{highlight_green(60/(20-c)=100/(20+c))}}}.


Simplify and solve for c.

--------
--------

Same type exercise comes up too often.
Something moves upstream for U distance in the same time as it moves D distance downstream.  The craft or vehicle normally moves at R speed in absence of stream or river current.
What is the speed of the current?

<pre>
              SPEED           TIME          DIST.

UPSTR         R-c             U/(R-c)         U

DOWNSTR       R+c             D/(R+c)         D
</pre>
{{{U/(R-c)=D/(R+c)}}}


{{{U(R+c)=D(R-c)}}}

{{{UR+Uc=DR-Dc}}}

{{{Uc+Dc=DR-UR}}}

{{{(U+D)c=R(D-U)}}}

{{{highlight_green(c=R((D-U)/(D+U)))}}}