On the opening night of a play at a local theatre, 895 tickets were sold for a total of $10,514. Adult tickets cost $14 each, children's tickets cost $11 each and senior citizen tickets cost $8 each. If the combined number of children and adult tickets exceeded twice the number of senior citizen tickets by 295, than how many tickets of each type were sold?

adults --- x

children -- y
seniors --- z

x+y+z = 895 , #1

14x +11y + 8z = 10514 , #2

translation of:
"If the combined number of children and adult tickets exceeded twice the number of senior citizen tickets by 295"
---> x+y > 2z by 295
x+y = 2z + 295
x + y - 2z = 295 #3

#1 - #3:
3z = 600
z = 200
so #1
---> x+y + 200 = 895
y= 695 - x

sub into #2:
14x + 11(695-x) +8(200) = 10514
14x + 7645 - 11x + 1600 = 10514
3x = 1269
x = 423
then y = 695 - 423 = 272

So they sold
423 adult
272 children and
200 senior tickets

checK
14(423) + 11(272) + 8(200) = 10514, YEAHHH

Ticket Sales. Stacey is selling tickets to the school play. The tickets are $7 for adults and $4 for children. She sells twice as many adult tickets as children's tickets and brings in a total of $306. How many of each kind of ticket did she sell?

Let's assume that the number of adult tickets sold is x, the number of children's tickets sold is y, and the number of senior citizen tickets sold is z.

1. From the given information, we know that:
- x + y + z = 895 (equation 1) (Total number of tickets sold)
- 14x + 11y + 8z = 10,514 (equation 2) (Total revenue from ticket sales)

2. We are also given that the combined number of children and adult tickets exceeded twice the number of senior citizen tickets by 295. Mathematically, this can be expressed as:
- x + y > 2z + 295 (equation 3)

Now, let's solve the equations to find the values of x, y, and z.

From equation 1, we can express x in terms of y and z:
x = 895 - y - z

Substituting this value of x into equations 2 and 3, we get:

14(895 - y - z) + 11y + 8z = 10,514
895 + 14y - 14z + 11y + 8z = 10,514
25y - 6z = 9,619 (equation 4)

895 - y - z + y > 2z + 295
895 - z > 3z + 295
605 > 4z
z < 605/4
z < 151.25

Since the number of senior citizen tickets must be a whole number, z can be 151 at the maximum.

Now we can find the possible values for x and y using equation 4.

Let's assume z = 151:
25y - 6(151) = 9,619
25y - 906 = 9,619
25y = 10,525
y = 421

Substituting the values of y and z into equation 1, we can find x:
x + 421 + 151 = 895
x = 323

Therefore, if z = 151, the number of adult tickets sold (x) is 323, the number of children's tickets sold (y) is 421, and the number of senior citizen tickets sold (z) is 151.

To solve this problem, let's break it down into equations.

Let's assume the number of adult tickets sold is "A", the number of children's tickets sold is "C", and the number of senior citizen tickets sold is "S".

1) We know that the total number of tickets sold is 895:
A + C + S = 895

2) We also know that the total revenue from ticket sales is $10,514:
14A + 11C + 8S = 10,514

3) The combined number of adult and children's tickets exceeds twice the number of senior citizen tickets by 295:
A + C > 2S + 295

Now, we have a system of three equations. To solve it, we can use a method called "substitution." Let's solve equation 1 for A and substitute it into equation 2 and equation 3.

From equation 1, we have A = 895 - C - S.

Now, substitute that into equation 2:
14(895 - C - S) + 11C + 8S = 10,514

Expanding this equation will give us:
12,530 - 14C - 14S + 11C + 8S = 10,514

Simplifying further will give us:
-3C - 6S = -1,016
3C + 6S = 1,016 (equation 4)

Now, substitute A = 895 - C - S into equation 3:
(895 - C - S) + C > 2S + 295

Simplifying this gives us:
895 - S > S + 295
2S < 600
S < 300

Now, we have all the information we need to solve the system of equations. We can plug the value of S into equation 4 to find C, and then substitute both S and C into equation 1 to find A.

Let's try S = 200:
3C + 6(200) = 1,016
3C + 1,200 = 1,016
3C = -184
C = -61.33 (This solution is not meaningful as the number of tickets must be positive)

Let's try S = 250:
3C + 6(250) = 1,016
3C + 1,500 = 1,016
3C = -484
C = -161.33 (Again, not a meaningful solution)

Let's try S = 280:
3C + 6(280) = 1,016
3C + 1,680 = 1,016
3C = -664
C = -221.33 (Once again, not a meaningful solution)

Let's try S = 290:
3C + 6(290) = 1,016
3C + 1,740 = 1,016
3C = -724
C = -241.33 (Still not meaningful)

Let's try S = 295:
3C + 6(295) = 1,016
3C + 1,770 = 1,016
3C = -754
C = -251.33 (This doesn't work either)

Based on this, we can see that using the given conditions, we cannot find positive values for the number of tickets sold for each type. It's possible that there may be an error in the problem's conditions or that there is no real solution. I advise double-checking the problem and its given conditions.