Let's define the variables for the number of tickets:
- Let \( x \) be the number of first-class tickets bought.
- Let \( y \) be the number of coach tickets bought.
We know from the problem statement that:
-
The total number of people (tickets) is 13: \[ x + y = 13 \]
-
The total cost for the tickets was $8,990: \[ 950x + 390y = 8990 \]
Now we can solve this system of equations.
First, let's solve the first equation for \( y \): \[ y = 13 - x \]
Next, we substitute this expression for \( y \) into the second equation: \[ 950x + 390(13 - x) = 8990 \]
Expanding the equation: \[ 950x + 5070 - 390x = 8990 \]
Combining like terms: \[ (950 - 390)x + 5070 = 8990 \] \[ 560x + 5070 = 8990 \]
Subtracting 5070 from both sides: \[ 560x = 8990 - 5070 \] \[ 560x = 3920 \]
Dividing both sides by 560: \[ x = \frac{3920}{560} = 7 \]
Now that we have \( x \), we can find \( y \): \[ y = 13 - x = 13 - 7 = 6 \]
So, Janine bought:
- \( 7 \) first-class tickets
- \( 6 \) coach tickets
Answer: First class tickets: 7; Coach tickets: 6.