4 men and 6boys can finish …

Let the man finishes the work in x days and that the boy finishes in y days.

One day's work of a man = {tex}\frac{1}{x}{/tex}

and one day's work of a boy ={tex}\frac{1}{y}{/tex}

Since, 4 men and 6 boys finish a piece of work in 5 days.

{tex}\therefore{/tex} One day's work of 4 men and 6 boys = {tex}\frac{1}{5}{/tex} of the work

{tex} \Rightarrow \frac{4}{x} + \frac{6}{y} = \frac{1}{5}{/tex}  

Similarly, in second case, 

One day's work of 3 men and 4 boys = {tex}\frac{1}{7}{/tex} part of the work

{tex}\therefore \frac{3}{x} + \frac{4}{y} = \frac{1}{7}{/tex}

Thus, we have the following equations

{tex}\frac{4}{x} + \frac{6}{y} = \frac{1}{5}{/tex} ......(i)

and

{tex}\frac{3}{x} + \frac{4}{y} = \frac{1}{7}{/tex} .....(ii)

Here , Eqs. (i) and (ii) are not in linear form , so we reduce them in linear form by putting {tex}\begin{array}{l}\frac1x\;=\;u\;and\;\frac1y\;=\;v\\\end{array}{/tex}

Now, Eq. (i) becomes {tex}4u + 6v = \frac{1}{5}{/tex}  .....(iii)

and Eq(ii) becomes {tex}3u + 4v = \frac{1}{7}{/tex} .....(iv)

On multiplying Eq(iii) by 3 and Eq(iv) by 4 and then subtract  Eq , we get

{tex}18v - 16v = \frac{3}{5} - \frac{4}{7}{/tex}

{tex}\Rightarrow 2v = \frac{{21 - 20}}{{35}} \Rightarrow 2v = \frac{1}{{35}} \Rightarrow v = \frac{1}{{70}}{/tex}

Put {tex}v=\frac{1}{70}{/tex} in Eq(iv), we get {tex}3u + \frac{4}{{70}} = \frac{1}{7}{/tex}

{tex} \Rightarrow 3u = \frac{1}{7} - \frac{4}{{70}}{/tex}

{tex} \Rightarrow 3u = \frac{6}{{70}}{/tex}

{tex} \Rightarrow u = \frac{1}{{35}}{/tex}

Thus {tex}u=\frac{1}{35}{/tex} and {tex}v=\frac{1}{70}{/tex}

{tex} \Rightarrow \frac{1}{x} = \frac{1}{{35}}{/tex}and {tex}\frac{1}{y} = \frac{1}{{70}}{/tex}

{tex} \Rightarrow x = 35{/tex} and {tex}y=70{/tex}

Hence, 1 man alone and 1 boy alone finishes the work in 35 and 70 days, respectively.