SOLUTION: A woman can bicycle 46 miles in the same time as it takes her to walk 6 miles. She can ride 10 mph faster than she can walk. How fast can she walk?

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Question 1033236: A woman can bicycle 46 miles in the same time as it takes her to walk 6 miles. She can ride 10 mph faster than she can walk. How fast can she walk?
Found 2 solutions by josgarithmetic, josmiceli:
Answer by josgarithmetic(39613) About Me  (Show Source):
You can put this solution on YOUR website!
The description seems in general to be familiar.
            RATE        TIME      DISTANCE


BICYCLE       r+10        t         46


WALK          r           t          6

Lookup Question number 1032831.
https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.1032831.html

That one only showed a data table and no further steps, as is being done here too. Can you form the necessary equations and solve the system?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = her speed walking in mi/hr
+s+%2B+10+ = her speed riding in mi/hr
Let +t+ = her time in hrs for both
walking and riding
-----------------------------------
Equation for riding bicycle:
(1) +46+=+%28+s+%2B+10+%29%2At+
equation for walking:
(2) +6+=+s%2At+
-----------------------
(2) +t+=+6%2Fs+
Plug this result into (1)
(1) +46+=+%28+s+%2B+10+%29%2A%286%2Fs%29+
(1) +46s+=+6s+%2B+60+
(1) +40s+=+60+
(1) +s+=+1.5+
She walks 1.5 mi/hr
---------------------
check:
(2) +6+=+s%2At+
(2) +6+=+1.5%2At+
(2) +t+=+4+ hrs
and
(1) +46+=+%28+s+%2B+10+%29%2At+
(1) +46+=+%28+1.5+%2B+10+%29%2At+
(1) +46+=+11.5t+
(1) +t+=+46%2F11.5+
(1) +t+=+4+ hrs
OK