8 men and 12 boys can …

Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days. Then,
one man's one day's work $= \frac { 1 } { x }$
One boy's one day's work $= \frac { 1 } { y }$
$\therefore$ Eight men's one day's work = $\frac { 8 } { x }$
$12\ boy's$ one day's work = $\frac { 12 } { y }$
According to question it is given that  $8\ men$ and $12\ boys$ can finish the work in $10\ days$
$10 \left( \frac { 8 } { x } + \frac { 12 } { y } \right) = 1 \Rightarrow \frac { 80 } { x } + \frac { 120 } { y } = 1$ .................(i)
Again, $6\ men$ and $8\ boys$ can finish the work in $14\ days$.
$\therefore \quad 14 \left( \frac { 6 } { x } + \frac { 8 } { y } \right) = 1 \Rightarrow \frac { 84 } { x } + \frac { 112 } { y } = 1$ ...........(ii)
Putting $\frac { 1 } { x } = u$ and $\frac { 1 } { y } = v$ in equations (i) and (ii), we get
$80u + 120u - 1 = 0$
$84u + 112v - 1 = 0$
By using cross-multiplication, 
$\Rightarrow \frac { u } { - 120 + 112 } = \frac { - v } { - 80 + 84 } = \frac { 1 } { 80 \times 112 - 120 \times 84 }$
$\Rightarrow \quad \frac { u } { - 8 } = \frac { v } { - 4 } = \frac { 1 } { - 1120 }$
$\Rightarrow \quad u = \frac { - 8 } { - 1120 } = \frac { 1 } { 140 } \text { and } v = \frac { - 4 } { - 1120 } = \frac { 1 } { 280 }$
$u = \frac { 1 } { 140 } \Rightarrow \frac { 1 } { x } = \frac { 1 } { 140 } \Rightarrow x = 140$
$v = \frac { 1 } { 280 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 280 } \Rightarrow y = 280$
One man alone can finish the work in $140\ days$ and one boy alone can finish the work in $280\ days$.