Question 1108941
Her rate returning was 3 mph LESS than the rate going,
so her rate going could not have been 1.19 mph.
 
{{{x}}}= Sue's rowing rate (speed) going (in mph) is what we are asked for.
{{{x-3}}}= Sue's rowing rate returning (in mph)
Each way was 7 miles, so it took Sue
{{{7/x}}} hours for the going across, and
{{{7/(x-3)}}} hours for the return trip.
The total time in hours was
{{{7/x+7/(x-3)=3}}}
We need to solve that equation for {{{x}}} .
 
{{{7/x+7/(x-3)=3}}}
{{{7(x-3)/(x(x-3))+7x/(x(x-3))=3}}}
{{{(7x-21)/(x(x-3))+7x/(x(x-3))=3}}}
{{{(7x-21+7x)/(x(x-3))=3}}}
{{{(14x-21)/(x^2-3x)=3}}}
{{{14x-21=3(x^2-3x)}}}
{{{14x-21=3x^2-9x}}}
{{{0=3x^2-9x-14x+21}}}
{{{3x^2-23x+21=0}}}
The solutions to that equation are {{{x=(23 +- sqrt(277))/6}}} ,
which round to approximately {{{6.6072}}} and {{{1.05945}}} .
However, {{{x=1.05945}}} would make {{{x-3<0}}} ,
and we cannot have a negative speed.
So, with a more reasonable number of significant digits,
we could say that Sue's rate going across was {{{highlight(6.6mph)}}} ,
which would make the rate {{{6.6mph-3mph=3.6mph}}} for the return trip.
That would have made the approximate rowing times
{{{7miles/"6.6 mph"}}}{{{hours=about}}}{{{1.06hours}}} going across the lake, and
{{{7miles/"3.6 mph"}}}{{{hours=about}}}{{{1.94hours}}} returning.