Question 692018
{{{b}}} = rate of the boat in calm water (in miles per hour)
{{{c}}} = rate of the current (in miles per hour)
 
With the current, the boat moves at {{{b+c}}} miles per hour.
In {{{3}}} hours it covers a distance (in miles) of
{{{3(b+c)=54}}}
 
Against the current, the boat moves at {{{b-c}}} miles per hour.
In {{{3.6}}} hours it covers a distance (in miles) of
{{{3.6(b-c)=54}}}
 
Those two equations form a system of linear equations that needs to be solved for {{{b}}} and {{{c}}},
but they can be simplified a bit.
{{{3(b+c)=54}}} --> {{{3(b+c)/3=54/3}}} --> {{{b+c=18}}}
{{{3.6(b-c)=54}}} --> {{{3.6(b-c)/3.6=54/3.6}}} --> {{{b-c=15}}}
Now we can write the system as
{{{system(b+c=18,b-c=15)}}}
 
You can easily solve the system by substitution, or by elimination.
 
BY ELIMINATION:
Adding both equations, we get
{{{2b=18+15}}} --> {{{2b=33}}} --> {{{2b/2=33/2}}} --> {{{highlight(b=16.5)}}}
Substituting that into {{{b+c=18}}, we get
{{{16.5+c=18}}} --> {{{16.5+c-16.5=18-16.5}}} --> {{{highlight(c=1.5)}}}
 
BY SUBSTITUTION:
{{{b-c=15}}} --> {{{b-c+c=15+c}}} --> {{{b=15+c}}}
Substituting {{{15+c}}} for {{{b}}} in {{{b+c=18}}}, we get
{{{15+c+c=18}}} --> {{{15+2c=18}}} --> {{{15+2c-18=18-15}}} --> {{{2c=3}}} --> {{{2c/2=3/2}}} --> {{{highlight(c=1.5)}}}
Substituting that into {{{b=15+c}}}, we get
{{{b=15+1.5}}} --> {{{highlight(b=16.5)}}}