Answer:
x = 8 ( 20$ bills)
y = 5 ( 10 $ bills)
z = 2 ( 5 $ bills)
Explanation:
Let call x, y, and z the number of bill of 20, 10, and 5 $ respectively
then according to problem statement, we can write
20*x + 10*y + 5*z = 220 (1)
We also know the total number of bills (15), then
x + y + z = 15 (2)
And that quantity of 20 $ bill is equal to
x = 3 + y (3)
Now we got a three equation system we have to solve for x, y, and z for which we can use any valid procedure.
As x = 3 + y by substitution in equation (2) and (1)
( 3 + y ) + y + z = 15 ⇒ 3 + 2*y + z = 15 ⇒ 2*y + z = 12
20* ( 3 + y ) + 10*y + 5*z = 220 ⇒ 60 + 20*y + 10*y + 5*z = 220
30*y + 5*z = 160 (a)
Now we have only 2 equations
2*y + z = 12 ⇒ z = 12 - 2*y
30*y + 5*z = 160 30*y + 5* ( 12 - 2*y) = 160
30*y + 60 - 10*y = 160
20*y = 100
y = 100/20 y = 5 Then by substitution in (a)
30*y + 5*z = 160
30*5 + 5*z = 160
150 + 5*z = 160 ⇒ 5*z = 10 z = 10/5 z = 2
And x
x + y + z = 15
x + 5 + 2 = 15
x = 8