algebra precalculus - Time and work . Men and women working.

It takes six days for three women and two men working together to
complete a work. Three men would do the same work five days sooner
than nine women. How many times does the output of a man exceed that
of a woman?

Question?

The statement "Three men would do the same work five days sooner
than nine women."

Does this mean that three men would complete "any amount" of the "given type of work" five days earlier than 9 women? By which I mean to say if there are 100 units of work, then three men would complete the work 5 days earlier than 9 women. Also if there are 1000 units of work, then also three men would complete the work 5 days earlier than 9 women.Is this the case?How?

If this is the case then we can write:

Let a man do $m$ units of work/day and a woman do $w$ units of work/day.
Let the time taken by 9 women to do the entire work be $t$ days.

$(3w+2m)*6=1$

$3m(t-5)=1$

$9w*t=1$

If this is not the case then we have:

$(3w+2m)*6=1$

$3m(t-5)=9w*t$

In this case where is the 3rd equation that we need to find the variables?

Answer

Let one woman complete the job in $w$ days $\implies$ rate of woman’s work is $\frac{1}{w}$ job/day.

Let one man complete the job in $m$ days $\implies$ rate of man’s work is $\frac{1}{m}$ job/day.

Your mistake is in the step: $(3w+2m)\times 6=1$. It should instead be $$(\frac{3}{w}+\frac{2}{m})\times 6 = 1$$

Now, as three men would do the work five days sooner than nine women, we have, $$\frac{m}{3}+5=\frac{w}{9}$$

Solving the quadratic, we have, $m=15$ and thus, $w=90$, giving us, $\frac{w}{m}=6$.