Question 1154228: adult tickets cost 8$ and student tickets for the concert cost 4$ exactly 290 tickets were sold and there was 1,440 made total how many students and adults tickets were sold
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website! .
From the condition, you have this system of 2 equations in 2 unknowns
x + y = 290 (1) (total number of tickets)
8x + 4y = 1440 (2) (total money)
where x = # of adult tickets, y = # of student tickets.
From equation (1), express y = 290-x and substitute it into equation (2), replacing y. You will get then
8x + 4*(290-x) = 1440.
From the last equation
x =  = 70.
ANSWER. 70 adult tickets and 290-70 = 220 student tickets.
CHECK. 8*70 + 4*220 = 1440 dollars. ! Correct !
Solved.
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It is a standard tickets problem.
There are different methods of solving such problems.
In this site, there are lessons
- Using systems of equations to solve problems on tickets
- Three methods for solving standard (typical) problems on tickets
explaining and showing all basic methods of solving such problems.
From these lessons, learn on how to solve such problems once and for all.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
Answer by greenestamps(13367)  (Show Source):
You can put this solution on YOUR website!
Formally....
a = number of adult tickets
s = number of student tickets
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)
Solve the pair of equations by your favorite method....
When the equations are both in that form, elimination is easiest.
4a+4s = 1160
8a+4s = 1440
subtract one equation from the other to eliminate s
4a = 280
a = 70
There were 70 adult tickets.
So there were 290-70 = 220 student tickets.
ANSWER: 70 adult tickets, 220 student tickets.
CHECK: 70(8)+220(4) = 560+880 = 1440
Informally -- if an algebraic solution is not required....
(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.
ANSWER: 70 adult tickets and 220 student tickets
Note that the informal solution used exactly the same calculations as the formal method....
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