Maya is signing her daughter up for summer activities. Some of the activities cost $10, and some of them cost $20. Maya pays a total of $130 for 9 activities. How many of each activity did she sign her daughter up for?

Let the number of activities costing $10 be $ x $, and the number of activities costing $20 be $ y $.

We can set up the following system of equations based on the problem:

1. The total number of activities is 9:
   \[
   x + y = 9
   \]

2. The total cost of the activities is $130:
   \[
   10x + 20y = 130
   \]

### Step 1: Simplify the second equation.
We can divide the second equation by 10 to make it easier to work with:
\[
x + 2y = 13
\]

### Step 2: Solve the system of equations.
Now, we have the system:
1. $ x + y = 9 $
2. $ x + 2y = 13 $

Subtract the first equation from the second:
\[
(x + 2y) - (x + y) = 13 - 9
\]
\[
y = 4
\]

### Step 3: Find $ x $.
Substitute $ y = 4 $ into the first equation:
\[
x + 4 = 9
\]
\[
x = 5
\]

### Final Answer:
Maya signed her daughter up for 5 activities costing $10 and 4 activities costing $20.

Would you like a detailed explanation of any of the steps? Here are some related questions you might find helpful:

1. What if the total number of activities was different?
2. How would you solve the problem if the prices of the activities were changed?
3. Can you generalize this method for any total cost and number of activities?
4. What would the solution look like if there were three types of activities with different costs?
5. What if some activities were free? How would that affect the equations?

**Tip**: When solving a system of equations, try to simplify one of the equations early on to make it easier to solve the system.

Learn how to solve a system of linear equations to determine how many activities costing $10 and $20 Maya signed her daughter up for, given that she paid a total of $130 for 9 activities. This problem helps students in Grades 6-8 understand key algebraic concepts.

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