Question 353479
As you stated they went 19 miles.
If you looking for the average speed then:
D=RT
19=(3/X+8/(19-X))4.25
19=(3[19-X]+8X)/X(19-X)4.25
19=(57-3X+8X)/(19X-X^2)4.25
19=(57+5X)/(19X-X^2)4.25
19=4.25(57+5X)/(19X-X^2)
19*(19X-X^2)=242.25+21.25X
361X-19X^2-242.25-21.25X=0
-19X^2+339.75X-242.25=0
19X^2-339.75X+242.25=0
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
X=(339.75+-SQRT[339.75^2-4*19*242,25])/2*19
X=(339.75+-SQRT[115,430.0625-18,411])/38
X=(339.75+-SQRT97,019.0625)/38
X=(339.75+-311.4788)/38
X=(339.75-311.4788)/38
X=28.2712/38
X=.744 MILES FOR THE 3 MPH TRIP.
19-.744=18.256 MILES FOR THE 6 MPH BOAT TRIP.
PROOF:
19=((3/.744)+8/(18.256))4,25
19=(4.032+.4382)4.25
19=4.47*4.25
19~19