On a windy day William found that he could travel 10 mi downstream and then 2 mi backupstream at top speed in a total of 16 min. What was the top speed of William's boat if the rateof the current was 30 mph? (Let x represent the rate of the boat in still water.)

Answer:The top speed of William's boat was 45 mphStep-by-step explanation:Letx -----> represent the rate of the boat in still water in mphwe know thatThe speed or rate is equal to divide the distance by the timespeed=distance /timetime=distance/speedDownstreamspeed=(30+x) mphdistance=10 mitime1=10/(30+x)Upstreamspeed=(x-30) mphdistance=2 mitime2=2/(x-30)The sum of the time downstream plus the time upstream must be equal to 16 minutesConvert minutes to hours[tex]16\ min=16/60\ h[/tex][tex]\frac{10}{x+30} +\frac{2}{x-30}=\frac{16}{60}[/tex]Multiply by (x+30)(x-30) both sides[tex]10(x-30)+2(x+30)=\frac{16}{60}(x^2-900)\\10x-300+2x+60=\frac{16}{60}x^2-240\\12x-240=\frac{16}{60}x^2-240\\\frac{16}{60}x^2-12x=0[/tex]Multiply by 60 both sides[tex]16x^2-720x=0[/tex]Divide by 16 both sides[tex]x^2-45x=0\\x(x-45)=0[/tex]The solution is x=45\ mph