SOLUTION: Mason & Joe walked to a dock at 3 miles per hour, got on a boat, and traveled, to dillion at 8 miles per hour. if the total distance was 19 miles and the trip took 4.25 miles in al

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Question 353479: Mason & Joe walked to a dock at 3 miles per hour, got on a boat, and traveled, to dillion at 8 miles per hour. if the total distance was 19 miles and the trip took 4.25 miles in all, how far did they go?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
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+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
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