SOLUTION: A bus company has 3,000 passengers daily, paying a $1.25 fare. For each 25¢ increase in fare, the company estimate that it will lose 80 passengers. What is the smallest increase in

Let x = the smallest number of 25¢ increases in the fare that will produce 
the $4970 daily revenue.

Then the new fare is $1.25 + $0.25x or 1.25 + 0.25x

For each of the x 25¢ increase in fare, the company will lose 80
passengers

Then the new number of passengers will be 3000 - 80x

Therefore the new daily revenue will be (3000 - 80x)(1.25 + 0.25x)

This must equal to $4970. So the equation is 

               (3000 - 80x)(1.25 + 0.25x) = 4970
Use FOIL:
               3750 + 750x - 100x - 20x² = 4970

                     -20x² + 650x - 1220 = 0

Divide thru by -10       2x² - 65x + 122 = 0

Factor:                     (2x-61)(x-2) = 0

                        2x-61=0;   x-2=0
                           2x=61;    x=2
                            x=30.5;  

The answer 30.5 makes no sense.  The only answer that makes sense is 
x = 2.  That is, 2 25¢ increases will bring in the desired revenue.

Checking:  The new fare will be $1.75. There will be 2×80 = 160 passengers
lost.  So there will be 3000-160 passengers or 2840 passengers, each paying
$1.75 so the total revenue will be 2840×$1.75 or $4970.  So the answer is
correct.

Edwin