Question 1113273
.
<pre>
1.  From the condition,

        {{{(2/3)*M}}} = {{{(3/4)*W}}}.     (1)


    Multiply both sides by 12. You will get

        8M = 9W.           (2)

    or

        M = {{{(9/8)*W}}}          (3)



2.  They ask you to evaluate the ratio of  {{{(2/3)*M}}} + {{{(3/4)*W}}}  (the amount of married people) to

    (M+W)  (the total adult population of the condo complex).  


    The numerator is  {{{(2/3)*M}}} + {{{(3/4)*W}}} = {{{(1/12)*(8M+9W)}}}   (4)


    In (4), replace 8M by 9W,  based on (2).  Then you can continue for the numerator

        {{{(2/3)*M}}} + {{{(3/4)*W}}} = {{{(1/12)*(8M+9W)}}} = {{{(1/12)*(9W + 9W)}}} = {{{(1/12)*18*W}}} = {{{(3/2)*W}}}.



     The denominator is  M + W = {{{(9/8)*W + W}}} = {{{(17/8)*W}}}.       ( <<<---=== I replaced  here M  by {{{(9/8)*W}}}  based on (3). )



3.  Now the ratio under the question

       {{{((2/3)*M + (3/4)*W)/(M+W)}}} = {{{((3/2)*W)/((17/8)*W)}}} = {{{((3/2))/((17/8))}}} = {{{(3*8)/(2*17)}}} = {{{(3*4)/17}}} = {{{12/17}}}.


<U>Answer</U>.  The ratio under the question = {{{12/17}}}.
</pre>

Solved.


=============


The technique should not cloud / (obscure) a simple idea behind the solution.


The most general formulation of such problems is THIS:


<pre>
    You are given two linear forms (functions) of two variable x and y:  p(x,y) = ax + by  and  q(x,y) = cx + dy.

    You are given also linear dependence between these two variables  {{{x/y}}} = e.

    Then they ask you evaluate the ratio  {{{(ax + by)/(cx+dy)}}}.


The solution is very simple:

    you express x = ey  and substitute it into each of the two linear functions:

        p(x,y) = ax + b*ex = (a + be)*x,

        q(x,y) = cx + d*ex = (c + de)*x.


Thus each of the two functions p(x,y)  and  q(x,y)  becomes the linear function of one variable (linear proportionality).


Then the ratio of the forms  p(x,y) = ax + by  and  q(x,y) = cx + dy  is

       {{{p(x,y)/q(x,y)}}} = {{{((a+b*e)x)/((c+d*e)*x)}}} = {{{(a+be)/(c + de)}}} = constant not depending on x and y.


That's all.
</pre>

So you actually can create/generate million of similar problems based on this technology . . . 



/\/\/\/\/\/\/\/\/


The post by the tutor &nbsp;@amalm &nbsp;is  &nbsp;&nbsp;<U>W R O N G</U>.


For your safety, &nbsp;simply ignore it . . .