Tickets for a certain show cost $14, $21, or, for VIP seats, $41. If ten times as many $14 tickets were sold as VIP tickets, and the number of $14 tickets sold was 54 more than the sum of the number of $21 tickets and VIP tickets, sales of all three kinds would total $48,446. How many of each kind of ticket would have been sold?

Number of $14 tickets sold = 1340

Number of $21 tickets sold = 1152

Number of VIP tickets sold = 134

Step-by-step explanation:

step 1 :

Let,

The equation is 14a + 21b + 41c = 48446     ----------(1)

Given,

Ten times as many $14 tickets were sold as VIP tickets.

⇒  10c = a  ---------(2)

step 2:

The number of $14 tickets sold was 54 more than the sum of the number of $21 tickets and VIP tickets.

⇒ a = 54 + (b+c)

⇒ a - 54 = b+c

Substitute a = 10c in the above eqn,

⇒ 10c - 54 = (b+c)

⇒ b = 9c - 54  -----------(3)

step 3 :

substitute (2) and (3) in eqn (1)

 14(10c) + 21(9c - 54) + 41c = 48446

Solve for the c value,

140c + 189c - 1134 + 41c = 48446

370c = 48446  + 1134

c = 49580 / 370

c = 134

step 4 :

Substitute c=134 in eqn (2),

10c = a

10 134 = a

a = 1340

Substitute c=134 in eqn (3),

b = 9(134) - 54

b = 1206 - 54

b = 1152

step 5 :

∴ a = number of $14 tickets = 1340

b = number of $21 tickets = 1152

c = number of VIP tickets = 134

it would take 5 hours i think idk i'm not that good at math.can you me with my questions i need asap.