# SOLUTION: I need help setting up this Linear Equation: While on vacation, Kevin went for a swim in a nearby lake. Swimming against the current, it took him 8 minutes to swim 200 meters.

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 Click here to see ALL problems on Linear-systems Question 581128: I need help setting up this Linear Equation: While on vacation, Kevin went for a swim in a nearby lake. Swimming against the current, it took him 8 minutes to swim 200 meters. Swimming back to shore with the current took half as long. Find Kevin's average swimming speed and the speed of the lake's current.Found 2 solutions by dfrazzetto, mananth:Answer by dfrazzetto(277)   (Show Source): You can put this solution on YOUR website! d=rt 200 = (r-c)8 200 = (r+c)4 r - c = 25 r + c = 50 ------------- 2r = 75 r = 37.5 c = 12.5 m/min average swimming speed = total distance / total time = 400/12 = 33 1/3 m/min Answer by mananth(12270)   (Show Source): You can put this solution on YOUR website!swimming speed =x mph current speed =y mph againstcurrent x-y 8 minutes with current x+y 4 minutes Distance = 200 m distance= 200 t=d/r against current 200.00 / ( x - y )= 8.00 8.00 x -8.00 y = 200.00 ....................1 200.00 / ( x + y )= 4.00 4.00 ( x + y ) = 200.00 4.00 x + 4.00 y = 200.00 ...............2 Multiply (1) by 1.00 Multiply (2) by 2.00 we get 8.00 x -8.00 y = 200.00 8.00 x + 8.00 y = 400.00 16.00 x = 600.00 / 16.00 x = 37.5 Meters /minute plug value of x in (1) 8 x -8 y = 200 300 -8 y = 200 -8 y = 200 -300 -8 y = -100 y = 12.5 Meters / minute swimming speed 37.5 Meters /minute (0.63m/s) current current 12.5 Meters /minute (0.21m/s)