Question 1203240
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There are $925 in $5, $10, and $20 bills in a cash register. There is a total of 71 bills. 
The number of $20 bills is 7 less than the total number of $5 bills and $10 bills. 
How many bills of each denomination are in the cash register?
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        I will show you how to solve the problem 
        using two unknowns (and two equations) only.



<pre>
Let x be the number of the  5-dollar bills;
Let y be the number of the 10-dollar bills.


Then the number of the 20-dollar bills is (x+y-7).


Write an equation for the total number of bills

    x + y + (x+y-7) = 71.


Simplify it

    2x + 2y = 71 + 7 = 78,

or

    x + y = 39.


Write second equation for the total money

    5x + 10y + 20(x+y-7) = 925.


Simplify it

     x + 2y + 4(x+y-7) = 185

     x + 2y + 4x + 4y - 28 = 185

    5x + 6y = 185 + 28 = 213


So, you have now the system of two equations

     x +  y =  39,    (1)

    5x + 6y = 213.    (2)


To solve it, multiply equation (1) by 5 (both sides); keep equation (2) as is.  You will get

    5x + 5y = 195,    (1')

    5x + 6y = 213.    (2')


Subtract equation (1') from equation (2').  You will get

    6y - 5y = 213 - 195

      y     =    18.


Now from equation (1)  x = 39-y = 39-18 = 21.


<U>ANSWER</U>.   There are 21 of the 5-dollar bills,  18 of the 10-dollar bills, and (x+y-7) = (21+18-7) = 32 of the 20-dollar bills.


<U>CHECK</U>.   21*5 + 18*10 + 32*20 = 925 dollars total.  ! correct !
</pre>

Solved.