Linear Systems

Question

Shirley walked a certain distance, then returned along the same route. She walked 10km/h going and returned at 8 km/h. How far did she walk if the route trip took 9 h?

Answer 1

Let x be the time the first part of the trip, and y be the time it took for the return part, both in hours.

 

Then since both parts of the trip are the same distance, one equation is 10x = 8y.

 

The other equation is for the total time, and is x + y = 9.

 

Isolate y in the second equation:

 

y = 9 - x

 

Plug it into the first equation:

 

10x = 8(9 - x)

 

10x = 72 - 8x

 

18x = 72

 

x = 4

 

We can plug this in to the first equation to get y = 5, but we don't need to for this problem.

 

Since the first part of the trip, Shirley walked for 4 hours at 10 km/h, the distance for each leg is

 

(4 h)(10 km/h) = 40 km.

 

The total distance of both parts of the trip is then 80 km.

Answer 2

Hello Kashaf,

 

Let's set this up with what we know

 

D = RT

distance = rate x time

 

R₁ = speed (rate) she walked going = 10 km/h

R₂ = speed (rate) she walked going home = 8 km/h

D₁ = distance she walked going

D₂ = distance she walked going home

D₁ = D₂

T₁ = time it took her going

T₂ = time it took her going home

T₁ + T₂ = 9

 

D₁ = R₁T₁

D₂ = R₂T₂

 

D₁ = 10T₁   and     T₁ = D₁/10

D₂ = 8T₂ and T₂ = D₂/8

 

Since T₁ + T₂ = 9

D₁/10 + D₂/8 = 9

 

Since D₁ = D₂, let's just replace both of them with D

So our equation becomes

D/10 + D/8 = 9

 

Now solve for D

(8D + 10D)/80 = 9

(8D + 10D) = 720

18D = 720

D = 40

 

She walked 40 km each way, 80 km round trip.