I don't understand what the rate and time is for the walk there and the ride back. 

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let t1 is the time needed for walking to Kates house, and t2 is the time needed for the ride back.

the distance covered for walking to Kates house and for riding back is the same.

so the distance for walking is 2 km/h * T1.   and the distance for riding back is 10 km/h * T2 it will be then

2 * T1  = 10 * T2  and we know that T1 + T2 = 3 hours  so. T1 = 3 - T2 substituting to first equation where

T1 = 3-T2   we have  2(3-T2) = 10*T2  and 6-2T2 = 10T2.  >>  6=10T2+2T2. ->>  6=12T2. ->>. T2=6/12

or T2= 1/6 hours. because T1+T2 = 3 hours T1= 3-T2. ->>. T1=3-1/6.  T1=17/6 hours
by Level 5 User (13.1k points)

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