31000 people attended a ballgames at a stadium that offers two kinds of seats; the day's receipts were $200000. how many people paid $14.00 for reserved seats, and how many paid$5.00 for general admission?

X= # of reserved seating
y= # of general admission

QUANTITY EQUATION:
x + y= 31,000

COST EQUATION:
14x + 5y= $200,000

STEP 1:
solve for either variable in quantity equation
x + y= 31,000
subtract both sides by y
x= 31000 -y

STEP 2:
substitute x=31000-y in cost equation
14x + 5y= $200,000

14(31000-y) +5y= 200000
434000 -14y + 5y= 200000

434000 - 9y= 200000
add 9y to both sides

434000= 200000 + 9y
subtract 200000 from both sides
234000= 9y

divide both sides by 9
26,000= y general admission

STEP 3:
substitute y answer in either equation
14x + 5y= $200,000
14x + 5(26000)= 200000
14x + 130000= 200000
14x= 70000
x= 5,000 reserved seats

CHECK:
14x + 5y= $200,000
14(5000) + 5(26000)= 200000
70000 + 130000= 200000
200000= 200000

ANSWER: There were 5,000 reserved seats and 26,000 general admission seats.

Hope this helps! :)

Ithink the right answer could be 50%