At 3: 30 p. m., berto’s train was 34 miles past the egg farm, traveling at an average speed of 85 miles per hour. at the same time on a nearby track, eduardo’s train was traveling at an average speed of 110 miles per hour and had 12 miles to go before it reached the egg farm. to the nearest hundredth of an hour, after how much time will the trains meet up?

E=egg farmx=distantce between  the Berto´s trains and the meeting point. t=time where the trains meet up ⇒110 miles/h                                          ⇒85miles/h[E]X[12 miles][34 miles][ x ]
distance between Berto´s train and Eduardo´s train==12 miles+34 miles=46 miles.

S=d/t      ⇒d=S*tS=seedd=distanceT=time
Eduardo´s train;46 miles+x=t*110 miles/h  ⇒x=t*110 miles/h-46 miles  (1)Berto´s train;x=t*85 miles/h  (2)
With the equations (1) and (2) we suggest this system equations:x=t* 110 miles/h-46 milesx=t*85 miles/hwe solve this system equiations :t*110 miles/h-46 miles=t*85 miles/h110t miles/h-85t miles/h=46 miles25 t miles/h=46 milest=46 miles/25 miles/h=1,84 hour  (≈1 hour, 50 minutes, 24 seconds)
x=t*85 miles/h=156,4 miles
Solution: 1,84 hour