It took a crew 80 minutes to row 3km upstream and back again. If the rate of flow of the stream was 3km/h, what was the rowing rate of the crew?

It is important to keep the units all the same.

As unit time for velocities is in hours:
Total time = 80 minutes `->80/60 hours`

Given that distance 1 way is 3Km

Let rowing velocity be `r`

Let time to row against current be `t_a`
Let time to row with current be `t_w`

Thus `t_w+t_a=80/60`

Known: distance is velocity x time
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Thus
For 'with current' `" "3Km = (r+3)t_w" "->" " t_w=3/(r+3)`
For against current`" "3Km=(r-3)t_a" "->" "t_a=3/(r-3)`

But `t_w+t_a=80/60`

`=>3/(r-3)+3/(r+3)=80/60`
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Consider that `a^2-b^2=(a+b)(a-b)`
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`=>(3(r+3)+3(r-3))/((r-3)(r+3) )" "->" "(6r)/(r^2-9)=80/60`

`=>(360r)/80=r^2-9`

`=>r^2-(360r)/80-9=0" note that "(360/80-=9/2)`
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Compare to `y=ax^2+bx+c" where "x=(-b+-sqrt(b^2-4ac))/(2a)`

`r=(-9/2+-sqrt(81/4-4(1)(-9)))/(2(1))`

`r=(-9/2+-sqrt(81/4+36))/(2)`

`r=(-9/2+-sqrt(225/4))/(2)`

`r=(-9/2+-sqrt(5^2xx7)/2)/2`

`r=-9/4+-(5sqrt(7))/4`

`=>r~~1.057" and "-3.077 " "(Km)/h`

The negative solution is not logical so

Rowing speed is:

`-9/4+(5sqrt(7))/4color(white)(..) (Km)/h` as an exact value

`1.057 color(white)(..)(Km)/h" "` ( to 3 decimal places) as an approximate value