Admission to a local aquarium is $10 for adults and $7 for children. The aquarium collected $6,459 from the sale of 783 tickets. How many children's tickets were sold?

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Answer 1 · 2016-12-12T13:23:29

There were 457 children tickets sold.

First, let's define our variables:

We can call the number of tickets for adults $a$ $a$

We can call the number of tickets for children $c$ $c$ .

Now we can right two equations which we can then solve using substitution:

$a + c = 783$ $a + c = 783$

$10 a + 7 c = 6459$ $10a + 7c = 6459$

Step 1) Solve the first equation for $a$ $a$ :

$a + c - c = 783 - c$ $a + c - c = 783 - c$

$a = 783 - c$ $a = 783 - c$

Step 2) Substitute $783 - c$ $783 - c$ for $a$ $a$ in the second equation and solve for $c$ $c$ :

$10 (783 - c) + 7 c = 6459$ $10(783 - c) + 7c = 6459$

$7830 - 10 c + 7 c = 6459$ $7830 - 10c + 7c = 6459$

$7830 - 3 c = 6459$ $7830 - 3c = 6459$

$7830 - 7830 - 3 c = 6459 - 7830$ $7830 - 7830 - 3c = 6459 - 7830$

$0 - 3 c = - 1371$ $0 - 3c = -1371$

$- 3 c = - 1371$ $-3c = -1371$

$\frac{- 3 c}{- 3} = \frac{- 1371}{- 3}$ $(-3c)/(-3) = (-1371)/(-3)$

$c = 457$ $c = 457$

Step 3) Substitute $457$ $457$ for $c$ $c$ for the solution to the equation in Step 1) and calculate $a$ $a$ :

$a = 783 - 457$ $a = 783 - 457$

$a = 326$ $a = 326$