Help me out in these linear equations

1. Sep 16, 2014

ajay.05

1. The problem statement, all variables and given/known data
2 women and 5 men can finish a work a work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 women alone to finish the work, and also that taken by 1 man alone.

2. Relevant equations
Pair of linear equations in two variables

3. The attempt at a solution
I can't figure how can I do it out:( LOL
But in the solution, I saw that it had been given that,
Work done by woman in 1 day=x
Work done by man in 1 day=y

=>2/x+ 5/y =1/4

=>3/x + 6/y =4/3

Solving them, gave the answers. But, my doubt here is, how did 2/x,5/y,etc. arrive?

2. Sep 16, 2014

Staff: Mentor

Strange, these are not linear equations.

Whole work is 1 - if a woman does x per day, she will need 1/x days to finish the work. 2 women need 2/x.

3. Sep 16, 2014

Ray Vickson

Why should 2 women need twice as long to finish the work?

4. Sep 16, 2014

Staff: Mentor

Good question, no idea what I was thinking

5. Sep 16, 2014

LCKurtz

Because they get yakity yaking??

6. Sep 16, 2014

Ray Vickson

Anyway, the RHS of the second equation above should be 1/3, not 4/3.

7. Sep 16, 2014

olivermsun

I think the confusion comes from the units in your problem set up.

Your setup says x is the work done by a woman in 1 day, so I take units are [work woman$^{-1}$ day$^{-1}$].
If so, then the work per day by 2 women would be: 2 [women] * x [work woman$^{-1}$ day$^{-1}$] = 2x [work day$^{-1}$].

Then the first equation would be 2x + 5y = 1/4 [work / day]. This is the way I'd do the problem.

Let x = the time taken by 1 woman alone to finish the work. Then x has units [woman days work$^{-1}$].
It follows that 2 [women] / x [woman days work$^{-1}$] has units [work days$^{-1}$], and you get the equations that you posted (with a 1/3 on the RHS as the previous poster points out).