An engaging Time & Work problem

Algebraic approach:

Let $x, y, and \ z$ denote the first man, the second man, and woman, respectively.

Then, translated to algebra:

$\begin{aligned} 3+ \frac{1}{1/x+1/z}=y \\\\ \frac{1}{1/y+1/z}=x \\\\ 2y-8=x \end{aligned}$

When we solve this system of equations, we get hours spent on the task, respectively. $x, y, and \ z$ spend $4, 6, and \ 12$ hours on the task, respectively.

In order to obtain the time needed to complete the task if they all work together, we need to solve:

$\frac{1}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} = \ ?$ , which yields $2$ when substituted.

$2 \ hours \ = \boxed{120}$ minutes.

$Let~ the~ number~ of~ hours ~to ~complete ~the~ task~ be~x,~y,~ z~~ respectively. \\ \\ Second ~man~ takes~ y~ hours,~ so~ other~ two ~need ~~y-3,~~ and~ equation~ is:-\\ \dfrac y y=(y-3)(\dfrac 1 x + \dfrac 1 z). ~~~~~~\implies~\dfrac 1 {y-3}=\dfrac 1 x+\dfrac 1 z.............(1) \\ \\ Second~man~and~women~too~take~x~hours, ~so~equation ~becomes~\dfrac x x =\dfrac x y+ \dfrac x z, ~reduces~to~\dfrac 1 x=\dfrac 1 y+\dfrac 1 z......(2)\\ ~~\\ First~man ~needs~x ~hours~second~man ~needs~h~hours.~So~x=2h-8.~~\implies~h=\frac 1 2 *(x+8).\\ So~\dfrac x x=\dfrac {x+8}{2y},~~~\implies~\dfrac 1 x=\dfrac 1{ 2y-8}.................(3)\\ ~~~\\ From~(1)~and~(2)\\ \dfrac 1 x =\dfrac 1 {y-3}-\dfrac1 z.\\ \dfrac 1 x =\dfrac 1 y+\dfrac1 z.\\ Adding~the~ two,\\ \dfrac 2 x= \dfrac 1 {y-3}+\dfrac 1 y.~~But~from~(3)\\ \dfrac 2 x= \dfrac 2{ 2y-8}.\\ \implies~\dfrac 1 {y-3}+\dfrac 1 y.=\dfrac 2{ 2y-8}.\\ \therefore~simplifying~y^2 - 8 y+12=0.\\ \\ Solving~this~quadratic~in ~y,~~y=2~or~6.\\ Since~x>0, ~from ~(3),~y=2~makes~x<0. ~So~y=2,~invalid.\\ \\ From~(3),~y=6~gives~x=4.\\ From~(2),~z=12.\\ \\ Let~H ~hours ~be ~the~ time~for~all~three~on~the~job.\\ So~\dfrac 1 x+\dfrac 1 y+\dfrac 1 z=\dfrac 1 H .\\ So~H=2~ hours=\huge \color{#D61F06}{120}~minutes.$