SOLUTION: cynthia bought five blank vhs tapes and four cassette tapes for $28.50. Charlie bought four vhs tapes and two cassette tapes for $21.00 how can I find the cost of each type of tap

Algebra ->  Linear-equations -> SOLUTION: cynthia bought five blank vhs tapes and four cassette tapes for $28.50. Charlie bought four vhs tapes and two cassette tapes for $21.00 how can I find the cost of each type of tap      Log On


   



Question 73035: cynthia bought five blank vhs tapes and four cassette tapes for $28.50.
Charlie bought four vhs tapes and two cassette tapes for $21.00 how can I find the cost of each type of tape?

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Let v=price of vhs tape, c=price of cassette tape
Set up a system of equation
                   
                       5v+4c=28.5
                       4v+2c=21                 

First you need to eliminate one variable. Multiply the entire 2nd equation by -12
                   
                       5v+4c=28.5
                    -2(4v+2c)=-2(21)                   

                   
                       5v+4c=28.5
                      -8v-4c-42                   

Now add each like term (eg 4c-4c=0c)
                   
                       -3v=-13.5
                        v=4.5                

So the price of a vhs tape is $4.50. Use this info to solve for c
                   
                       4(4.5)+2c=21                 
                       18+2c=21
                       2c=3
                       c=3/2=1.5                              

So the price of a vhs tape is $4.50 and the price of a cassette tape is $1.50.
Check:
                   
                       5(4.5)+4(1.5)=28.5
                       4(4.5)+2(1.5)=21                 

                   
                       22.5+6=28.5
                       18+3=21                 

                   
                       28.5=28.5
                       21=21                 

Solution works for the entire system.