Question 173842
Generally, I select my variables for word problems so that what I am solving for is the required answer.  But in this case, it seems more straightforward to solve for the number of skirts and blouses in inventory at the start.


So:  {{{s}}} is the beginning number of skirts, and {{{b}}} is the beginning number of blouses.


We know that the value of all the skirts at the beginning must be {{{45s}}} and the value of all the blouses must be {{{35s}}}.  Furthermore we know that the value of the entire inventory is $51,750.


So we can write:  {{{45s+35b=51750}}}   (Eq. 1)


The value of the one-half of the skirts that sold must be {{{(45/2)s}}} and the value of the two-thirds of the blouses that sold must be {{{(2/3)(35b)}}} and the sum of these two expressions must be the total revenue, or $30,600.


Now we can write:  {{{(45/2)s+(70/3)b=30600}}}  (Eq. 2)


Multiply Equation 2 by -6:


{{{-135s-140b=-183600}}}  (Eq. 3)


Multiply Equation 1 by 4:


{{{180s+140b=207000}}}  (Eq. 4)


Add Equation 3 and Equation 4, term by term:


{{{-135s+180s-140b+140b=-183600+207000}}}


{{{45s=23400}}}


{{{s=520}}}


Take this value for {{{s}}} and substitute into Equation 1:


{{{45(520)+35b=51750}}}


{{{35b=28350}}}


{{{b=810}}}


So, the store started with 520 skirts, but sold half of them, leaving 260 in stock.  Also, the store started with 810 blouses, but sold two-thirds of them, leaving one-third in stock {{{810/3=270}}}.