# linear equation problem solving

• Jul 4th 2008, 12:41 AM
white
linear equation problem solving
Need help on these questions..

question 1- show that 2/c = 1/a+1/b

attempt:
2ab=bc + ac...

question 2 - Xiu travels from town A to B at u km/h and then returns at v km/h. Town A is d km from town B
a) if i t takes T hours to travel from A to B, find the time taken:
i) the entire trip in terms of T, u and v

attempt:
time = distance/ speed
u + v =speed back and forth
distance is d km
so d/u+v ??

question 3- A man walks 70 km. He walks x km at 8 km/h and y km at 10km/h
a) fins his average speed in terms of x and y.

attempt:
average speed is = distance/ time
70/8x+10y ??

thanks!
• Jul 4th 2008, 01:11 AM
Isomorphism
Quote:

Originally Posted by white
Need help on these questions..

question 1- show that 2/c = 1/a+1/b

attempt:
2ab=bc + ac...

How can we prove 2/c = 1/a+1/b without knowing anything about c,a and b?! So please post the complete question...

Quote:

question 2 - Xiu travels from town A to B at u km/h and then returns at v km/h. Town A is d km from town B
a) if it takes T hours to travel from A to B, find the time taken:
i) the entire trip in terms of T, u and v

attempt:
time = distance/ speed
u + v =speed back and forth
distance is d km
so d/u+v ??
Wrong...

(a) says "it takes T hours to travel from A to B".So there is no "back and forth" trip.

d km is the distance from A to B.

time = distance/ speed

$\displaystyle T = \frac{d}{u}$

Let the "back and forth trip" take time 't', then the total time is:

$\displaystyle t = \frac{d}{u} + \frac{d}{v}$

But $\displaystyle d = uT$, so:

$\displaystyle t = \frac{uT}{u} + \frac{uT}{v} = T\left(1 + \frac{u}{v}\right)$

Thus $\displaystyle t = T\left(1 + \frac{u}{v}\right)$ is the time taken for the entire trip in terms of T, u and v.

Quote:

question 3- A man walks 70 km. He walks x km at 8 km/h and y km at 10km/h
a) fins his average speed in terms of x and y.

attempt:
average speed is = distance/ time
70/8x+10y ??
average speed = total distance/total time

$\displaystyle \text{total distance }= x + y = 70$

$\displaystyle \text{total time }= \frac{x}{8} + \frac{y}{10}$
• Jul 4th 2008, 10:16 PM
fresh_
question 1
a man on a bicycle rides one third of the way from town A to town B at a speed a km/h and the remainder of the way at 2blm/h

show that 2/c = 1/a+1/b
• Jul 4th 2008, 10:34 PM
Isomorphism
Quote:

Originally Posted by fresh_
question 1
a man on a bicycle rides one third of the way from town A to town B at a speed a km/h and the remainder of the way at 2blm/h

show that 2/c = 1/a+1/b

Ok but what is c?
• Jul 4th 2008, 10:41 PM
white
sorry guys , i didnt email the full quessiton to fresh .
here it is..
question 1
a man on a bicycle rides one third of the way from town A to town B at a speed a km/h and the remainder of the way at 2blm/h
a) if the distance between the two towns is 9km find the time taken to ride form A to B.

if the man travelled at a uniform rate of 3c km/h, he could jave ridden from A to B and back again in the same time

b)show that 2/c = 1/a+1/b
• Jul 4th 2008, 11:12 PM
Isomorphism
Quote:

Originally Posted by white
sorry guys , i didnt email the full quessiton to fresh .
here it is..
question 1
a man on a bicycle rides one third of the way from town A to town B at a speed a km/h and the remainder of the way at 2blm/h
a) if the distance between the two towns is 9km find the time taken to ride form A to B.

if the man travelled at a uniform rate of 3c km/h, he could jave ridden from A to B and back again in the same time

b)show that 2/c = 1/a+1/b

Since the total distance is 9 km, one third is 3 km. Thus the total time taken to move from A to B is:

$\displaystyle T = \frac{3}{a}+\frac{6}{2b}$

So to move from A to B and back again, he travels 18 km at a uniform rate of 3c km/hr, the time he takes is $\displaystyle \frac{18}{3c}$.

According to the question, this is same as T.

So

$\displaystyle T =\frac{18}{3c} = \frac{3}{a}+\frac{6}{2b}$

$\displaystyle \frac{6}{c} = \frac{3}{a}+\frac{3}{b}$

$\displaystyle \frac{2}{c} = \frac{1}{a}+\frac{1}{b}$