SOLUTION: Assume that a current exchange rate ,the British pound is worth 1.65 in American dollars.you have some dollar bills and several British pound coins.there are 17 items all together,

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Question 1204055: Assume that a current exchange rate ,the British pound is worth 1.65 in American dollars.you have some dollar bills and several British pound coins.there are 17 items all together,which have a value of 21.55 in American dollars.how many American dollars and how many British pound coins do you have?
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!

Assume that a current exchange rate ,the British pound is worth 1.65 in American dollars.
you have some dollar bills and several British pound coins.
Let dollar bills be x
and pound coins be y
there are 17 items all together
x+y=17................................................(1)
,which have a value of 21.55 in American dollars.
x+1.65y = 21.55...................................(2)
subtract (1) from (2)
x+1.65y = 21.55
x+y=17
0.65y = 4.55
y= 4.55/0.65
y=7
x+y=17
x+7 =17
x=10
you have 10 American dollars and 7 British pound coins.
Check
10+7*1.65=21.55





Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Assume that a current exchange rate ,the British pound is worth 1.65 in American dollars.
you have some dollar bills and several British pound coins.
there are 17 items all together, which have a value of 21.55 in American dollars.
how many American dollars and how many British pound coins do you have?
~~~~~~~~~~~~~~~~~ 

Let x be the number of British pound coins.
Then the number of American dollar bills is (17-x).


Write the value equation in American dollars

    1.65x + (17-x) = 21.55  dollars.


Simplify and find x

    1.65x + 17 - x = 21.55

    1.65x - x = 21.55 - 17

       0.65x  =    4.55

           x  =    4.55/0.65 = 7.


ANSWER.  There are 7 British pound coins and the rest, 17-7 = 10, are American one-dollar bills.

Solved.