Joseph sold tickets to the school musical. He had 12 bills worth $175 for the ticketssold. If all the money was $5 bills, $10 bills, and $20 bills, how many of each bill did he have?

if he had x,y,z bills of $5,$10,$20 respectively,

x+y+z = 12
5x+10y+20z = 175

A third equation would help, so we have to do some trial and error.

1 $5
5 $10
6 $20

To solve this problem, we can use a system of equations. Let's represent the number of $5 bills, $10 bills, and $20 bills Joseph had as x, y, and z, respectively.

From the information given, we know that Joseph had a total of 12 bills, so our first equation is:

x + y + z = 12

We also know that the total value of the bills is $175, so our second equation is:

5x + 10y + 20z = 175

Now we have two equations that we can solve simultaneously to find the values of x, y, and z.

We can start by simplifying the second equation by dividing both sides by 5:

x + 2y + 4z = 35

Now we have a simpler equation that we can use to substitute into the first equation.

Using substitution, we can rewrite the first equation as:

x + y + z = 12

Rearranging the simplified second equation, we can solve for x:

x = 35 - 2y - 4z

Now we substitute this expression for x into the first equation:

35 - 2y - 4z + y + z = 12

Combine like terms:

- 2y - 4z + y + z = 12 - 35

Simplify:

- y - 3z = -23

Now, we can express y in terms of z:

y = 23 - 3z

Finally, substitute these expressions for x and y into the equation x + y + z = 12:

(35 - 2y - 4z) + y + z = 12

Combine like terms:

35 - 2y - 4z + y + z = 12

Simplify:

-y - 3z = -23

Substitute y = 23 - 3z:

-(23 - 3z) - 3z = -23

Simplify:

-23 + 3z - 3z = -23

-23 = -23

The equation simplifies to a true statement, which means there are infinitely many solutions to this problem.

Therefore, there are multiple combinations of x, y, and z that satisfy the given conditions. We cannot determine a unique solution without additional information.