SOLUTION: I am trying to help my son with his algebra and need some help! I have two I am stuck on can you please help. I need an equation to model and solve. Below are the questions.

Question 485610: I am trying to help my son with his algebra and need some help! I have two I am stuck on can you please help. I need an equation to model and solve. Below are the questions.
1) Attendance at a ball game was 400 people. Student tickets cost $2 and adult tickets $3. $1050 was collected in ticket sales, how many of each were sold.
2) Find two consecutive integers such that the sum of the first and three times the second is 55.
Thank you,
Dwayne
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
1) Attendance at a ball game was 400 people. Student tickets cost $2 and adult tickets $3. $1050 was collected in ticket sales, how many of each were sold.
----------------
s + a = 400
2s + 3a = 1050
---------------
2s + 2a = 800 1st eqn times 2
2s + 3a = 1050
-------------- subtract
-a = -250
a = 250 adult tix
s = 150 student tix
---------------------
2) Find two consecutive integers such that the sum of the first and three times the second is 55.
n + 3(n+1) = 55
n + 3n + 3 = 55
4n + 3 = 55
4n = 52
n = 13
--> 13 & 14

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
400 people
student costs $2.00 apiece
Adult costs $3.000 apiece.
total collected was $1050.
x = number of student tickets
y = number of adult tickets.
x + y = 400
2x + 3y = 1050
solve simultaneously.
multiply first equation by 2 to get:
2x + 2y = 800
2x + 3y = 1050
subtract first equation from second equation to get:
y = 250
since x + y = 400, then x = 150.
2*150 + 3*250 = 300 + 750 = 1050.
number of child tickets is 150
number of adult tickets is 250

integers are x and x+1
sum of first and 3 times second is 55.
x + 3(x+1) = 55
x + 3x + 3 = 55
4x + 3 = 55
4x = 52
x = 13
x+1 = 14
13 + 3 * 14 = 13 + 42 = 55.

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