linear equation problem solving: show 2/c = 1/a+1/b, etc

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yumm

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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! :D
 
Re: linear equation problem solving

Hello, yumm!

Could you repost #1? .As written, it doesn't make sense . . .

2) Xiu travels from town \(\displaystyle A\) to \(\displaystyle B\) at \(\displaystyle u\) km/hr, then returns at \(\displaystyle v\) km/hr.
. . Town \(\displaystyle A\) is \(\displaystyle d\) km from town \(\displaystyle B.\)

a) If it takes \(\displaystyle T\) hours to travel from \(\displaystyle A\) to \(\displaystyle B\),
. . find the time taken: for the entire trip in terms of \(\displaystyle T,\:u\text{ and }v.\)
Click to expand...

\(\displaystyle \text{Time}\: = \:\frac{\text{Distance}}{\text{Speed}}\)

\(\displaystyle \text{From }A\text{ to }B:\;\text{ distance}= d\text{ km, }\:\text{speed}= u\text{ km/hr.}\)
. \(\displaystyle \text{This takes: }\:\frac{d}{u}\text{ hours.}\)
\(\displaystyle \text{Since this time is }T\text{ hours: }\:\frac{d}{u} \:=\:T\quad\Rightarrow\quad d \:=\:Tu\;\;\bf[1]}\)

\(\displaystyle \text{From }B\text{ to }A:\:\text{distance} = d\text{ km, }\;\text{speed} = v\text{ km/hr.}\)
. \(\displaystyle \text{This takes: }\:\frac{d}{v}\text{ hours.}\)
\(\displaystyle \text{Substitute \bf[1]}: }\:\frac{Tu}{v}\text{ hours.}\)

\(\displaystyle \text{Therefore: }\:\text{total time} \:=\:T + \frac{Tu}{v}\,\text{ hours.}\)



3- A man walks 70 km. He walks \(\displaystyle x\) km at 8 km/hr and \(\displaystyle y\) km at 10 km/hr.
a) Find his average speed in terms of \(\displaystyle x\) and \(\displaystyle y.\)
Click to expand...

\(\displaystyle \text{Important: }\:\text{Average speed} \:=\:\frac{\text{Total distance}}{\text{Total time}}\)


\(\displaystyle \text{We are given: }\:\text{Total distance} = 70\text{ km}\)


\(\displaystyle \text{He walked }x\text{ km at 8 km/hr.}\)
. . \(\displaystyle \text{This took: }\:\frac{x}{8}\text{ hours.}\)

\(\displaystyle \text{He walked }y\text{ km at 10 km/hr.}\)
. . \(\displaystyle \text{This took: }\:\frac{y}{10}\text{ hours.}\)
\(\displaystyle \text{His total time is: }\:\frac{x}{8} + \frac{y}{10}\,\text{ hours.}\)


\(\displaystyle \text{Therefore: }\:\text{Average speed}\:=\:\frac{70}{\frac{x}{8} + \frac{y}{10}} \text{ km/hr}\)

 
here
question 1 - Show that 2/c = 1/a + 1/b

so its 2 over c = 1 over a plus 1 over b (fraction)

same ?
 
Re: linear equation problem solving

soroban said:
Hello, yumm!

Could you repost #1? .As written, it doesn't make sense . . .

2) Xiu travels from town \(\displaystyle A\) to \(\displaystyle B\) at \(\displaystyle u\) km/hr, then returns at \(\displaystyle v\) km/hr.
. . Town \(\displaystyle A\) is \(\displaystyle d\) km from town \(\displaystyle B.\)

a) If it takes \(\displaystyle T\) hours to travel from \(\displaystyle A\) to \(\displaystyle B\),
. . find the time taken: for the entire trip in terms of \(\displaystyle T,\:u\text{ and }v.\)
Click to expand...

\(\displaystyle \text{Time}\: = \:\frac{\text{Distance}}{\text{Speed}}\)

\(\displaystyle \text{From }A\text{ to }B:\;\text{ distance}= d\text{ km, }\:\text{speed}= u\text{ km/hr.}\)
. \(\displaystyle \text{This takes: }\:\frac{d}{u}\text{ hours.}\)
\(\displaystyle \text{Since this time is }T\text{ hours: }\:\frac{d}{u} \:=\:T\quad\Rightarrow\quad d \:=\:Tu\;\;\bf[1]}\)

\(\displaystyle \text{From }B\text{ to }A:\:\text{distance} = d\text{ km, }\;\text{speed} = v\text{ km/hr.}\)
. \(\displaystyle \text{This takes: }\:\frac{d}{v}\text{ hours.}\)
\(\displaystyle \text{Substitute \bf[1]}: }\:\frac{Tu}{v}\text{ hours.}\)

\(\displaystyle \text{Therefore: }\:\text{total time} \:=\:T + \frac{Tu}{v}\,\text{ hours.}\)



[quote:129agf0k]3- A man walks 70 km. He walks \(\displaystyle x\) km at 8 km/hr and \(\displaystyle y\) km at 10 km/hr.
a) Find his average speed in terms of \(\displaystyle x\) and \(\displaystyle y.\)
Click to expand...

\(\displaystyle \text{Important: }\:\text{Average speed} \:=\:\frac{\text{Total distance}}{\text{Total time}}\)


\(\displaystyle \text{We are given: }\:\text{Total distance} = 70\text{ km}\)


\(\displaystyle \text{He walked }x\text{ km at 8 km/hr.}\)
. . \(\displaystyle \text{This took: }\:\frac{x}{8}\text{ hours.}\)

\(\displaystyle \text{He walked }y\text{ km at 10 km/hr.}\)
. . \(\displaystyle \text{This took: }\:\frac{y}{10}\text{ hours.}\)
\(\displaystyle \text{His total time is: }\:\frac{x}{8} + \frac{y}{10}\,\text{ hours.}\)


\(\displaystyle \text{Therefore: }\:\text{Average speed}\:=\:\frac{70}{\frac{x}{8} + \frac{y}{10}} \text{ km/hr}\)

[/quote:129agf0k]
for question 3 the answer actually 80(x+y) / 10x+8y ( form back of the textbook answer)
and q.2 is vT +uT/ v
 
yumm said:
here: question 1 - Show that 2/c = 1/a + 1/b
Click to expand...
Given only the information provided (just the equation, with no information regarding the variables or any relationships), there is no way to "show" this, because it simply isn't true. For instance, if a = 2, b = 3, and c = 4, then 2/c = 1/2, but 1/a + 1/b = 1/2 + 1/3 = 5/6.

Please reply with the full text of the exercise, including the context and instructions. Thank you.

soroban said:
\(\displaystyle \text{...2)Therefore: }\:\text{total time} \:=\:T + \frac{Tu}{v}\,\text{ hours.}\)

\(\displaystyle \text{...3) Therefore: }\:\text{Average speed}\:=\:\frac{70}{\frac{x}{8} + \frac{y}{10}} \text{ km/hr}\)
Click to expand...
yumm said:
for question 3 the answer actually 80(x+y) / 10x+8y ( form back of the textbook answer) and q.2 is vT +uT/ v
Click to expand...
In case this poster doesn't reply immediately, giving you the rest of your homework, you might want to spend a couple of minutes working on the exercises yourself: simplify the answers you were given, rearranging them into the book's equivalent forms on your own.

One is often surprised by how much more a student generally learns and grows when he is enabled to be more than just a copyist.

Eliz.
 
1- Show that 2/c = 1/a + 1/b

Are you certain you meant 2/c as opposed to 1/c.

Assuming you are looking for positive integer solutions:

Since a, b, and c are positive integers, a > c and b > c. Letting a = c + u and b = c + v, with u and v > 0, the equation 1/a + 1/b = 1/c reduces to c^2 = uv. Therefore, for every c, we need only decompose c^2 into the product of two integers u and v and solve for a, b and c.

I have yet to have any success with 2/c.
 
Re:

stapel said:
yumm said:
here: question 1 - Show that 2/c = 1/a + 1/b
Click to expand...
Given only the information provided (just the equation, with no information regarding the variables or any relationships), there is no way to "show" this, because it simply isn't true. For instance, if a = 2, b = 3, and c = 4, then 2/c = 1/2, but 1/a + 1/b = 1/2 + 1/3 = 5/6.

Please reply with the full text of the exercise, including the context and instructions. Thank you.

soroban said:
\(\displaystyle \text{...2)Therefore: }\:\text{total time} \:=\:T + \frac{Tu}{v}\,\text{ hours.}\)

\(\displaystyle \text{...3) Therefore: }\:\text{Average speed}\:=\:\frac{70}{\frac{x}{8} + \frac{y}{10}} \text{ km/hr}\)
Click to expand...
yumm said:
for question 3 the answer actually 80(x+y) / 10x+8y ( form back of the textbook answer) and q.2 is vT +uT/ v
Click to expand...
In case this poster doesn't reply immediately, giving you the rest of your homework, you might want to spend a couple of minutes working on the exercises yourself: simplify the answers you were given, rearranging them into the book's equivalent forms on your own.

One is often surprised by how much more a student generally learns and grows when he is enabled to be more than just a copyist.

Eliz.
Click to expand...

yerh i understand question 2 , how i got to it, but question 1...
70/ x/8 + y/10= i know how they got the bottom denominater but i dont get how they got the top
80(x+y) / 10x+8y
 
yumm said:
yerh i understand question 2 , how i got to it, but question 1...
70/ x/8 + y/10= i know how they got the bottom denominater but i dont get how they got the top
80(x+y) / 10x+8y
Click to expand...
I'm going to guess that you mean that you've figured out (or been given) the rest of the solution to exercise (2), but have no idea how to get started on the last step of exercise (3).

To learn how to simplify complex fractions, try studying some online lessons:

. . . . .Google results for "simplifying complex fractions"

Once you have studied at least two lessons from the above link, please attempt the last step of the solution process. If you get stuck, please reply showing some work of your own. Thank you.

Eliz.
 
70/ x/8 + y/10 .

common denominater is 80,
so
5600/80 /10x/80 + 8y/80

so 5600/80 divided by 10x/80 + 8y/80 = 5600/80 multiply 80/10x+8y= 5600/10x + 8y T__T :x :x failed again
 
yumm said:
70/ x/8 + y/10 .

common denominater is 80,
so
5600/80 /10x/80 + 8y/80

so 5600/80 divided by 10x/80 + 8y/80 = 5600/80 multiply 80/10x+8y= 5600/10x + 8y T__T :x :x failed again
Click to expand...
yumm, start by posting PROPERLY; that should be 70 / (x/8 + y/10) : brackets are REQUIRED.

Your answer 5600/10x + 8y should be shown as: 5600 /(10x + 8y); it is correct:
remember that 70 = x + y; so: 80(x + y) / (10x + 8y)
 
Question 1- show that 2/c = 1/a+1/b

If 2/c is what you meant:

1--1/a + 1/b = 2/c
2--Let a = c + u and b = c + v
3--1/(c + u) + 1/(c + v) = 2/c
4--Expanding and simplifying yields c = -2uv/(u + v)
5--Solutions are few and far between.
6--For example, u = 3 and v = -1, c = 3, a = 6 and b = 2
7--1/6 + 1/2 = 2/3 or 1/6 + 3/6 = 2/3 or 4/6 = 2/3
8--Similarly, for u = 6 and v = -2; c = 6, a = 12 and b = 4
9--1/12 + 1/4 = 2/6 or 1/12 + 3/12 = 4/12 = 2/6
 
Question 1
2/C = 1/a + 1/b

find common denominator(cab) : 2ab/cab = bc/cab + ac/cab

since there is a common denominator the numerators have to equal
2ab = bc + ac or 2ab= c(a+b)

Solve for C
C= 2ab / (a+b)

Plug into original eqn

2 / (2ab/(a+b)) = 1/a +1/b
(mult. by reciprical) 2 * (a+b) / 2ab = 1/a +1/b (find common denominator of right side (ab))
(2's cancel) (a+b) /ab = b /ab + a/ab
(a+b) /ab = (a+b) /ab all done!!!! :lol:


if someone wants to right this is a better eqn form it might help.
 
stapel said:
Given only the information provided (just the equation, with no information regarding the variables or any relationships), there is no way to "show" this, because it simply isn't true. For instance, if a = 2, b = 3, and c = 4, then 2/c = 1/2, but 1/a + 1/b = 1/2 + 1/3 = 5/6.
Click to expand...
JUST BECAUSE IT DOES NOT WORK FOR CERTAIN NUMBERS DOES NOT MEAN IT IS NOT TRUE I JUST POSTED THE ANSWER AT THE BOTTOM CHECK IT OUT - CROW626
 
crow626 said:
JUST BECAUSE IT DOES NOT WORK FOR CERTAIN NUMBERS DOES NOT MEAN IT IS NOT TRUE I JUST POSTED THE ANSWER AT THE BOTTOM CHECK IT OUT - CROW626
Click to expand...
Actually, the proof that an equation is not always true is exactly as was shown: the display of a counter-example. So, as posted, the equation that the poster is supposed to "prove" to be true is in fact false. Unless there is some additional information which has yet to be provided, there is nothing to "prove", because the statement, under the current conditions, is known to be false.

What you did was rearrange the equation. This is a restatement of it; it is not a proof of validity.

Eliz.
 
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