Answer:

Step-by-step explanation:
Cost of first 4 kilometers ride = 11 pesos.
So, for the first 4 km, the fair is constant i.e.
for 1 km the fare is 11 pesos,
for 2 km the fair is also 11 pesos,
similarly, for 3 km of 4 km, the fare is 11 pesos.
Hence, for the distance
, the fair function,

After 4 km, there is an increment of $ 3.50 for each kilometer.
So, the fare function up to 5 kilometers,

So, the fare function up to 6 kilometers, i.e for the distance
,

This can be arranged as the fare function up to 6 km,

Similarly, the fare function up to
kilometer
,


Hence, from equations (i) and (ii),
