Respuesta :
You can set up a system of equations for this problem. x= number of coach tickets and y = number of first class tickets.
$210x + $1200y = $10,230 (cost of coach ticket plus cost of first class tickets is total budget)
x + y = 11 (number of coach tickets plus number of first class tickets is total number of people)
Solve the second equation for y to get y = 11 - x, then plug that into the first equation and solve for x:
$210x + $1200(11 - x) = $10,230
$210x + $13,200 - $1200x = $10,230
-$990x + $13,200 = $10,230
-$990x = $2,970
x = 3
Sarah bought x = 3 coach tickets. Plug that into the second equation and solve for y:
3 + y = 11
y = 8
Sarah bought y = 8 first class tickets.
$210x + $1200y = $10,230 (cost of coach ticket plus cost of first class tickets is total budget)
x + y = 11 (number of coach tickets plus number of first class tickets is total number of people)
Solve the second equation for y to get y = 11 - x, then plug that into the first equation and solve for x:
$210x + $1200(11 - x) = $10,230
$210x + $13,200 - $1200x = $10,230
-$990x + $13,200 = $10,230
-$990x = $2,970
x = 3
Sarah bought x = 3 coach tickets. Plug that into the second equation and solve for y:
3 + y = 11
y = 8
Sarah bought y = 8 first class tickets.
The number of coach tickets purchased is 3, and the number of first-class tickets purchased is 8.
Let us assume the number of coach tickets purchased is x.
And the number of first-class tickets purchased is y.
Now,
Including Sarah, a total of 11 people took the trip.
Thus,
[tex]x+y=11[/tex] ............(1)
The cost of coach tickets and first-class tickets are $210 and $1200 respectively.
The total budget for airfare for the trip is $10,230.
Therefore,
[tex]210x+1200y=10230\\21x+120y=1023[/tex]............(2)
Solving both the equations, we get.
[tex]x=3\\y=8[/tex]
Thus, The number of coach tickets purchased is 3, and the number of first-class tickets purchased is 8.
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https://brainly.com/question/7480814