Trip distance calculation

In three days, the students went 65 km on a trip. On the first day, they ran twice as much as on the third day, and on the second day, they ran 10 km less than on the first day. How many kilometers did they cover each day?

Final Answer:

a =  30
b =  20
c =  15

Step-by-step explanation:


a+b+c=65
a = 2c
b = a-10

a+b+c=65
a = 2·c
b = a-10

a+b+c = 65
a-2c = 0
a-b = 10

Row 2 - Row 1 → Row 2
a+b+c = 65
-b-3c = -65
a-b = 10

Row 3 - Row 1 → Row 3
a+b+c = 65
-b-3c = -65
-2b-c = -55

Pivot: Row 2 ↔ Row 3
a+b+c = 65
-2b-c = -55
-b-3c = -65

Row 3 - -1/-2 · Row 2 → Row 3
a+b+c = 65
-2b-c = -55
-2.5c = -37.5


c = -37.5/-2.5 = 15
b = -55+c/-2 = -55+15/-2 = 20
a = 65-b-c = 65-20-15 = 30

a = 30
b = 20
c = 15

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