**METHOD 1:**

*Strategy: Temporarily disregard the first 1/5 mile.*Dan pays $2.00 for the first fifth mile and $7.60 for the rest of the ride. Each fifth of a mile costs 40¢, and 760 ÷ 40 = 19. The $7.60 pays for a ride of 19 fifths of a mile. Adding 1/5 mile for the first $2.00,

**the longest ride Dan can afford is 20/5 or 4 miles.**

**METHOD 2:**

*Strategy: Make a table.*The cost increases by $.40 for each fifth of a mile. When the desired amount is reached, the taxi will have traveled a length of 20/5 or 4 miles.

**METHOD 3:**

*Strategy: Use algebra.*Let N = the number of fifths of a mile after the first fifth mile that Dan can ride.

2.00 + .40N = 9.60

.40N = 7.60 (Subtract 2.00 from each side)

N = 19 (Divide each side by .40)

Add the first fifth of a mile, bringing the length of the ride to 20/5 or 4 miles.