Use the given information to write a system of equations. Isolate one of the variables and substitute the obtained expression into the other equation.
Price in June: $60 Price in July: $50
Practice makes perfect
First, let's call the price of the DVD set in June x and the number of sold sets in June y. From the exercise we know that in July the price was $10 less and there were 30 more sets sold.
Price in June:& x
Price in July:& x-10
Sets Sold in June:& y
Sets Sold in July:& y+30
We also know that MediaWorld had revenue of $2700 in June and $3750 in July. Knowing that the revenue is the product of the price of a set and the number of sold sets, we get the following system of equations.
xy=2700 & (I) (x-10)(y+30)=3750 & (II)Notice that we are only interested in solving this system for x. Therefore, we will start by isolating y from the first equation and then substituting the obtained expression for y in the second equation. Let's do it!
Note that we have x in the denominator. Therefore, x cannot be 0. Since in our case x represents a price, 0 would not make sense, so we can proceed to solving. Let's substitute y into the second equation.
Notice that in Equation (II) we have a quadratic equation in terms of only the x-variable.
x^2 -45x - 900 = 0 ⇔ 1x^2+( -45)x+( -900)=0
We can substitute a= 1, b= -45, and c= -900 into the Quadratic Formula.
This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.
x=45 ± 75/2
x_1=45 + 75/2
x_2=45 - 75/2
x_1=120/2
x_2=- 30/2
x_1=60
x_2=- 15
Note that in our case negative price does not make any sense, so x=60 is the only possible solution. Therefore, the price of the DVD set in June was $ 60. We can now determine the price in July.
Price in July: 60-10=50
The price of the DVD set in July was $50.
