Question 1376
Regrettably, Khwang's answer is wrong:

"Hence, his speed going to the fair was 4 miles/hr,and 
his speed backing home was 8 miles/hr."

The question states that Antoine's total travel time was 3 hours.   If he travelled even 1 hour at 4 miles/h or 8 miles/h he would exceed 3 hours of travelling time.

"Antoine rode his bike the 4 miles to the fair at a relaxed pace. "

trip to the fair:
4 miles @ a speed of x mph

"He stayed to long and had to double his speed on the way back in order to get home in time for dinner."

trip home:
4 miles @ a speed of 2x mph

The question states that the entire trip took three hours, so we will build an equation with = 3 hours on the right hand side.

Now we include the entire trip on the Left hand side of the equation.   Since we're solving for time we flip the miles/h ratio to h/miles, so that the unit "miles" cancels and we're left with hours:  We have an algebraic representation of the speed of the trip above.

4 miles * h/(x miles)       +  4 miles * h/(2x miles)          = 3 hours
         
miles cancels out:

4h/x + 4h/2x = 3h


Get the x out of the denominators by multiplying every term by 2x:

8h + 4h = 6hx

12h = 6hx
 
divide through by 6h

2 = x

Now we know the value of x, plug it in to the speed ratios above and you have your answer.