SOLUTION: Jake works at Musicloud. He is responsible for checking the incoming shipments. An order of 20 portable radios and 25 CD players arrives, but the invoice is torn, so only the final

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Jake works at Musicloud. He is responsible for checking the incoming shipments. An order of 20 portable radios and 25 CD players arrives, but the invoice is torn, so only the final      Log On


   

Question 1149995: Jake works at Musicloud. He is responsible for checking the incoming shipments. An order of 20 portable radios and 25 CD players arrives, but the invoice is torn, so only the final charge of $1692.50 is readable. He remembers that that the CD players were $20 more than the radios. How much did one radio and one CD player cost? [Systems of Equations]
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let x be the cost of one CD player and y be the cost of one portable radio.


Then from the condition you have this system of 2 equations

    25x + 20y = 1692.50    (1)    (total cost)

      x -   y =   20       (2)


To solve it, multiply equation (2) by 20;  keep equation (2) as is.  Then you have

    25x + 20y = 1692.50    (1')  

    20x - 20y =  400      (2')


Next, add equations (1') and (2').  Then the terms "20y" will cancel each other, and you will get a single equation for x

    25x + 20x = 1692.50 + 400

    45x       = 2092.50

      x       = 2092.50%2F45 = 46.50.


ANSWER.  $46.50 for one CD player and 46.50 - 20 = 26.50 dollars for one portable radio.


CHECK.  46.50*25 + 26.50*20 = 1692.50  dollars.    ! Precisely correct ! !

Solved.