Gwen has $20, $10, and $5 bills in her purse worth a total of $220. She has 15 bills in all. There are 3 more $20 bills than there are $10 bills. How many of each does she have?

x = 8 (20$ bills)

y = 5 (10 $ bills)

z = 2 (5 $ bills)

Step-by-step 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