Please help: At a play 903 tickets were sold. The total receipts for the performance was $6,751.......

Child tickets cost $7 apiece, adult tickets were $9 apiece, and senior tickets were $5 apiece. If the combined number of child and adult tickets exceeded twice the number of senior tickets by 261, how many tickets of each type were sold?

Child:
Adult:
Senior:

c + a + s = 903 combining tickets
c + a = 2s + 261 a relation between the tickets
7c + 9a + 5s = 6751 relative prices of tickets

I'm going to take the first two equations and subtract them.
c-c = 0, a-a = 0, s-(-2s) = 3s, 903 - 261 = 642, therefore s = 214.
from eq.1 c + a + 214 = 903, therefore c = -a + 689
I'm going to plug s = 214 and c = -a + 689 into eq. 3
7(-a + 689) + 9a + 5(214) = 6751
-7a + 9a = 858, therefore a = 429
plug these two values into any of the 3 equations to find c.
c = -429 + 689 = 260

Child: 260
Adult: 429
Senior: 214

Suppose there were c child tickets, a adult tickets, and s senior tickets,
then you are given
c+a+s=903 (eqn1)
7c+9a+5s=6751 (eqn2)
c+a=2s+261 (eqn3)

Using (eqn1) and (eqn3), we get
2s+261+s=903
i.e., 3s=903-261=642
or s=642/3=214

Using this in (eqn3),
c+a=428+261=689 (eqn4)
Aslo using c=214 in (eqn2)
7c+9a+5*214=6751
7c+9a=6751-1070=5681 (eqn5)
So we have two simulatenous equations in c and a in (eqn4) and (eqn5) which we solve:
7 times (eqn4) is 7c+7a=689*7=4823 (eqn6)
Subtract (eqn6) from (eqn5)
2a=5681-4823=858
a=858/2=429

So from (eqn4) c+429=689
c=689-429=260

Thus there were 260 child tickets, 429 adult tickets, and 214 senior tickets sold.

You can check that they satisfy the given conditions.

Let a=# adult tickets, c=# child tickets, s=# senior tickets
so:
903=a+c+s and a+c=2*s+261

Hence 903=261+3*s so s=(903-261)/3 =642/3=214
and a+c=903-214=689

Now adding up to money we get: