Question 1154228
<br>
Formally....<br>
a = number of adult tickets
s = number of student tickets<br>
a+s = 290  (the total number of tickets was 290)
8a+4s = 1440  (the total cost, at $8 per adult and $4 per student, was $1440)<br>
Solve the pair of equations by your favorite method....<br>
When the equations are both in that form, elimination is easiest.<br>
4a+4s = 1160
8a+4s = 1440<br>
subtract one equation from the other to eliminate s<br>
4a = 280
a = 70<br>
There were 70 adult tickets.<br>
So there were 290-70 = 220 student tickets.<br>
ANSWER: 70 adult tickets, 220 student tickets.<br>
CHECK: 70(8)+220(4) = 560+880 = 1440<br>
Informally -- if an algebraic solution is not required....<br>
(1) If all 290 tickets had been student tickets, the total cost would have been $290*4 = $1160; the actual total is $1440-$1160 = $280 more than that.
(2) The difference between the cost of an adult ticket and a student ticket is $4.
(3) The number of adult tickets needed to make up that extra $280 is 280/4 = 70.<br>
ANSWER: 70 adult tickets and 220 student tickets<br>
Note that the informal solution used exactly the same calculations as the formal method....<br>