tickets cost $16, $25, $39. 10 times as many $16 tickets were sold as $39, number of $16 tickets sold was 54 more than $25 and $39 tickets, sales of all $55,042

Question

I'm stuck on this problem for my algebra homework.  I cannot get the correct equations to use to solve.  The question is asking for how many of each type of  ticket sold.  Any help is appreciated!

Answer 1

Let's start by listing our variables. We'll call the number of $16 tickets sold c for cheap, $25 tickets will be m for middle, and $39 tickets will be e for expensive.

Here's what we know:

c = 10e (ten times the amount of e sold were c)

c = m + e + 54

16c + 25m + 39e = 55042

 

Let's try to substitute a bit. We can convert the c = m + e + 54 into an isolated m so that we can replace it:

m = c - e - 54

 

16(10e) + 25(c - e - 54) + 39e = 55042

16(10e) + 25(10e - e - 54) + 39e = 55042

16(10e) + 25(9e - 54) + 39e = 55042

160e + 225e - 1350 + 39e = 55042

 

424e - 1350 = 55042

424e = 56392

e = 133

 

Now that we have e, we can solve for the other tickets:

c = 10 * 133

c = 1330

 

m = 1330 - 133 - 54

m = 1143

 

Now let's check our answer:

16*1330 + 25*1143 + 39*133 = 55042

21280 + 28575 + 5187 = 55042

49855 + 5187 = 55042

55042 = 55042