2/3 of the men are married to 3/4 of the women in Fractionville. What fraction of the people in Fractionville are married?
I've gotten it to 2/3x = 3/4y but I'm not sure what to do from here.

- anonymous

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- anonymous

you should have two equations, since you have two unknowns. so if you let x+y=1 you should be in good shape. (x and y are the corresponding fractions of the town that are male and female)

- phi

you need another equation
x+y=1
means the men and women add up to the total population

- anonymous

x = the number of men in fractionville and
y = the number of women in fractionville

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## More answers

- anonymous

how can x + y = 1 though?

- anonymous

if x is the fraction of the population that is male and y is the fraction of the town that is female, they should add up to 1 which is the fraction of people in the population

- anonymous

but how is x the fraction of- okay I kinda see, but explain how i get my x and y definitions?

- asnaseer

@JunkieJim is correct

- anonymous

i just dont understand it though shouldnt
x + y = P (population)?

- anonymous

Yes, that's exactly what we're doing, only we're dividing everything by P so you've got \[\frac{x}{Population} +\frac{y}{Population} = \frac{Population}{Population}\] so that everything remains as fractions

- phi

Think of it this way: some fraction of the people are men and some fraction are women
so x+y=1
the fraction of the population that are married is 2/3 x (another fraction)

- phi

which of course is the same fraction as 3/4 y

- anonymous

I appreciate how you are taking the time to explain it to me. And I'm understanding it better. I can do the systems from here, but that x + y = 1 is bothering me

- anonymous

what you said makes sense Phi, I'm just wondering how i got my definitions for x and y

- asnaseer

@mridrik: another way of looking at this is to call the total population p and the number of men m.
then, number of women = p - m
and you can write:\[\frac{2m}{3}=\frac{3(p-m)}{4}\]

- phi

your definitions should be modified
x = the fraction of the population that are men in fractionville and
y = the fraction of the population that are women in fractionville

- anonymous

so are my definitions wrong, or are they right but do not help me advance with the problem?

- phi

I would *not* use
x = the number of men in fractionville and
y = the number of women in fractionville

- anonymous

with asnaseers way, we get m = 9/17p, so how could I use that to get the answer?

- asnaseer

2m/3 are married

- phi

you can only find m/p (a fraction)

- anonymous

so 2/3 * 9/17 gives us ?

- anonymous

now I'm confused again

- phi

the fraction married

- anonymous

but we already knew the fraction married 2/3

- anonymous

i have a feeling i sound pretty stupid right now to you guys

- asnaseer

no - 2/3 is the fraction of men that are married - not the fraction of the population that are married.

- phi

double it (we have to count the women)

- anonymous

so 2/3 * 9/17 * 2?

- phi

Sounds good. we know 2/3 * 9/17 = 6/17 of the population is married men
3/4 * 8/17 is 6/17 of the population = married women
total fraction 12/17

- anonymous

You guys are genii! (I think thats genius plural)
Thank you, so 12/17 of the population of fractionville is married?

- asnaseer

:) we've just had a bit more practice than you have mridrik thats all. and yes phi has the correct answer.

- anonymous

It makes me happy to know that people care about another's will to learn.

- asnaseer

thats what makes this site so wonderful :)

- anonymous

If i could, i would make a program that gave each of you over 9000 medals

- asnaseer

:O) - careful - we may drown in glory - lol!

- anonymous

yeah lol, my math teacher couldnt even get this today in class

- asnaseer

try and be gentle on him - after all - he is also human :)

- anonymous

yes, maybe I'll get the chance to explain what you showed me to the class, even though i dont that well with public speaking, only about 20 students in my algebra 2 class

- asnaseer

good luck my friend...

- anonymous

thanks for your help, have you ever heard of mathcounts?

- asnaseer

no - what are they?

- anonymous

It's a program that my school participtes in with schools from around the United States, we go to regionals first, (district is too small) and usually win that, and then go to state where this one girl always wins, its from 6th to 8th grade (my last year) and its a blast when we go there to lexington this year (KY) but they questions get ridiculously hard towards the end of the competition and i bet you would beast them up

- asnaseer

I'm sure as time goes by you will become a mathematical ninja and win that contest hands down :)
I'm probably far too old for that now :)

- anonymous

oh no way, the way you win is the top ten scorers in the written test get to go to a quick recall type thing, and she has won that in 6th grade and 7th grade, I got 40 something place and my friend got in the 30s, but there are a LOT of people there so thats pretty good i guess, our school got 10th place overall last year

- asnaseer

that is an extremely good achievement. and never think that you will not be able to achieve even better goals for yourself - nothing in this world is impossible if you work hard enough at it. you certainly sound like someone who is willing to work hard to achieve high goals in life.

- anonymous

Thanks, I would like to be an engineer when I grow up, not really sure what they do, but everyone says it involves math so that sounds fun.

- asnaseer

yes - there is a lot of maths involved in engineering. your main focus should be to do something you really enjoy - after all, you will most likely be working for the rest of your life in one particular job - so might as well pick something you really enjoy.
anyway - must go now. good talking to you.

- anonymous

you too

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