<PRE><B>Question 290323</B>
Sue rowed her boat across and back lake bend in 3 hours.
 if her returning rate was 2mph less than the going rate, and if the distance each way was 7 miles find her rate going.
:
Yes, it could be:
{{{7/x}}} + {{{7/((x-2))}}} = 3 
Multiply by x(x-2), results
7(x-2) + 7x = 3x(x-2)
:
7x - 14 + 7x = 3x^2 - 6x
:
14x - 14 = 3x^2 - 6x
combine like terms to have a quadratic equation
0 = 3x^2 - 6x - 14x + 14 
:
3x^2 - 20x + 14 = 0
:
the quadratic formula 
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this problem a=3; b=-20; c=14
{{{x = (-(-20) +- sqrt(-20^2-4*3*14 ))/(2*3) }}}
:
{{{x = (20 +- sqrt(400-168 ))/6 }}}
:
{{{x = (20 +- sqrt(232 ))/6 }}}
Two solutions
{{{x = (20 + 15.23)/6 }}}
x = {{{35.23/6}}}
x = 5.87
and
{{{x = (20 - 15.23)/6 }}}
x = {{{4.77/6}}}
x = .795
:
The reasonable solution: x = 5.87 mph, her rate going
:
:
Check solution by finding the times coming and going
7/5.87 = 1.19, going
7/3.87 = 1.81, returning
----------------
total = 3 hrs

 </PRE>