Question 112443
I'm confused in your analysis, what is x and what is y?
For these types of problems, start with the following. 
Rate x Time = Distance
In your problem, the distance is constant at 20 miles. 
When the boat travels downstream, its total rate,{{{R[d]}}} is equal to the rate in still water,{{{R[b]}}}, plus the river rate{{{R[r]}}}.
So the equation downstream is 
Rate x Time = Distance
{{{R[d] (1)=20}}}
{{{(R[b]+R[r])(1)=20}}}
{{{R[b]+R[r]=20}}}
When the boat travels upstream, its total rate,{{{R[u]}}} is equal to the rate in still water,{{{R[b]}}}, minus the river rate{{{R[r]}}}.
So the equation downstream is 
Rate x Time = Distance
{{{R[u] (2.5)=20}}}
{{{(R[b]-R[r])(2.5)=20}}}
{{{R[b]-R[r]=8}}} Divide both sides by 2.5.
1.{{{R[b]+R[r]=20}}}
2.{{{R[b]-R[r]=8}}}
Adding equation 1 to equation 2 yields,
{{{2R[b]=28}}}
{{{R[b]=14}}}
and from equation 1,
1.{{{R[b]+R[r]=20}}}
{{{14+R[r]=20}}}
{{{R[r]=6}}}
The rate of the boat in still water is 14 mph. 
The river moves at a rate of 6 mph. 
Hope it helps.