# System of Linear Equations - Two Variable - Distance Time Speed Question

• Dec 16th 2010, 11:37 AM
who192
System of Linear Equations - Two Variable - Distance Time Speed Question
hey im stumped at developing a formula for this question:
Chris walks at 8 km/h and runs at 12 km/h. One day he walks and runs on the way from his house to the library. It takes him 20 minutes. On his way back from the library he runs twice as far and the journey home takes 17.5 minutes. How far is his house from the library?
Could you please show me how you did the question?
• Dec 16th 2010, 11:46 AM
snowtea
Let D be the total distance to the library.
Let d be the distance covered running *to* the library.
This means D - d is the distance covered walking *to* the library.

From the problem you know d/(12 km/h) + (D-d)/(8 km/h) = 20 min

The distance covered running *from* the library is 2d.
This means D - 2d is the distance covered walking *from* the library.

From the problem you know 2d/(12 km/h) + (D - 2d)/(8 km/h) = 17.5 min

Solve for D
• Dec 16th 2010, 12:00 PM
who192
Quote:

Originally Posted by snowtea
Let D be the total distance to the library.
Let d be the distance covered running *to* the library.
This means D - d is the distance covered walking *to* the library.

From the problem you know d/(12 km/h) + (D-d)/(8 km/h) = 20 min

The distance covered running *from* the library is 2d.
This means D - 2d is the distance covered walking *from* the library.

From the problem you know 2d/(12 km/h) + (D - 2d)/(8 km/h) = 17.5 min

Solve for D

thanks so much
when I did it as you told me to i got D=180 and the answer in my math book said 3 km
so then i did the same thing but i changed 20 min into 1/3 hour and 17.5 min into 7/24 hour and I got 3 km for D
most of the time these units mess me up
• Dec 16th 2010, 12:05 PM
snowtea
Yes, you always have to make sure your units match, which is why I included them.
Good job working out the correct answer.
• Dec 16th 2010, 02:22 PM
Soroban
Hello, who192!

Another approach . . .

Quote:

$\text}Chris walks at 8 km/hr and runs at 12 km/hr. \;One day he walks}$
$\text{and runs from his house to the library. \;It takes him 20 minutes.}$

$\text{On his way back, he runs twice as far and the journey home}$
$\text{takes 17.5 minutes. \;How far is his house from the library?}$

Going to the library, let $\,R$ = distance he ran, $\,W$ = distance he walked.

He ran $\,R$ km at 12 km/hr. .This took $\frac{R}{12}$ hours.

He walked $\,W$ km at 8 km/hr. .This took $\frac{W}{8}$ hours.

His total time is: . $20\text{ minutes }=\:\frac{1}{3}\text{ hour.}$

. . Then: . $\displaystyle \frac{R}{12} + \frac{W}{8} \:=\:\frac{1}{3} \quad\Rightarrow\quad 2R + 3W \:=\:8$ .[1]

Going home, he ran $2R$ km and walked $W-R$ km.

He ran $2R$ km at 12 km/hr. .This took $\frac{2R}{12}$ hours.

He walked $W-R$ km at 8 km/hr. .This took $\frac{W-R}{8}$ hours.

His total time is: . $17.5\text{ minutes }=\:\frac{7}{24}\text{ hour.}$

. . Then: . $\dfrac{2R}{12} + \dfrac{W-R}{8} \:=\:\dfrac{7}{24} \quad\Rightarrow\qud R + 3W \:=\:7$ .[2]

Subtract [1] - [2]: . $R \:=\:1 \quad\Rightarrow\quad W \:=\:2$

The total distance is: . $R + W \:=\:1 + 2 \:=\:3\text{ km.}$