Let's use x to represent the number of coach tickets that Sarah purchased, and y to represent the number of first class tickets. Since a total of 15 people took the trip, we know that:
x + y + 1 = 15
where the extra 1 represents Sarah.
We also know that Sarah used her entire budget of $8130 for airfare, so we can write an equation for the cost of the tickets:
220x + 910y = 8130
Now we have a system of two equations with two variables. We can solve for x and y using substitution or elimination.
Using substitution, we can solve for x in terms of y from the first equation:
x + y + 1 = 15 => x = 14 - y
Substituting this expression for x into the second equation, we get:
220(14 - y) + 910y = 8130
3080 - 220y + 910y = 8130
690y = 5050
y = 7.319 (rounded to three decimal places)
Since we cannot purchase a fractional number of tickets, we need to round y to the nearest whole number. Since y represents the number of first class tickets, Sarah must have purchased either 7 or 8 first class tickets.
If she purchased 7 first class tickets, then the number of coach tickets would be:
x = 14 - y = 14 - 7 = 7
This would result in a total cost of:
220(7) + 910(7) = 6440
But since Sarah's budget was $8130, she must have purchased more first class tickets than this.
If she purchased 8 first class tickets, then the number of coach tickets would be:
x = 14 - y = 14 - 8 = 6
This would result in a total cost of:
220(6) + 910(8) = 7280
This is less than Sarah's budget of $8130, so it is a possible solution.
Therefore, Sarah purchased 8 first class tickets and 6 coach tickets.