math problem help

Question

A store sells shirts for $45 and blouses for $35. Its entire stock is worth $51750. Yet, sales are slow and only half the shirts and two-thirds of the blouses are sold, for a total of $30600. How many shirts and blouses are left in the store?

Answer 1

Let x = number of shirts in stock

Let y = number of blouses in stock

 

Since shirts are $45, 45x represents the portion of the original stock, for shirts

Since blouses are $35, 35y represents the portion of the original stock, for blouses

 

    45x + 35y = 51750   

 

We can simplify the above by dividing both sides by 5

 

    9x + 7y = 10350   (Equation 1)

 

Since 1/2 of the shirts, and 2/3 of the blouses were sold,

 

    (1/2)(45x) + (2/3)35y = 30600 

 

We can clear the fractions in equation above by multiplying both sides of the equation by 6, the LCM of 2 and 3

 

     6(1/2)45x + 6(2/3)35y = 6(30600)

     135x + 140y = 183600

 

Simplify by dividing both sides by 5

 

     27x + 28y = 36600     (Equation 2)

 

Now we have our system of equations, simplified

 

      9x + 7y  =   10350      (Equation 1)

    27x + 28y = 36720       (Equation 2)

 

Multiply equation 1 by -3, then adding the resulting equations, (eliminating x)

 

    -27x - 21y = -31050

     27x +28y =  36720

     -------------------------------

                 7y = 5670

                   y = 810

 

Take equation 1, substitute 810 for y, then solve for x

 

     9x + 7y = 10350

     9x + 7(810) = 10350

     9x + 5670 = 10350

     9x = 10350 - 5670

     9x = 4680

     x = 520

 

Thus, there are 520 shirts and 810 blouses in stock