Question 686343: how do you slove this: aldult tickets for the school musical sold for $3.50 and student tickets sold for $2.50. 321 tickets were sold altogether for $937.50. how many of each kind of ticket was sold?
Answer by RedemptiveMath(80)   (Show Source):
You can put this solution on YOUR website! These types of questions involve methods to solve for systems of linear equations. It would prove well to first find the system of equations we are dealing with. That is, we need to find at least two different linear equations in order to solve for them. One of the best things to do is list each equation under a certain idea or theme. In this case, the information tells us we are dealing with tickets (or the number of people) and price. Since these are the two prominent ideas of this information, we can write it out in two equations:
a + s = 321
$3.50a + $2.50s = $937.50
You can choose what two variables you want to use as long as you keep them in constant understanding throughout your work (or if your teacher prefers you use certain variables). In this matter, I used variables a and s to represent adult tickets and student tickets respectively. I knew how to choose my variables because of the information we need to know. We need to know how many adult and student tickets were sold, but we do not. So, we use variables to stand in place for the missing information. Then we use manipulation to get what the variables equal, thus providing us with the information we need to know.
I knew how to write the first equation simply because the information told me that the amount of tickets sold is 321. There were only two parties who bought tickets: adults and students. So, this basically means that between adults and students, 321 tickets were sold, or a + s = 321. The only other piece of information that is important is price. Simply put, I wrote the sum of the amount of money for each adult ticket and student ticket, and I set that equal to the amount of money made from all of the tickets. $3.50 for each adult ticket can simply be written as $3.50a (or $3.50 times the amount of adult tickets sold), and $2.50s is the same as saying $2.50 for each student ticket sold.
Now, we can solve using three methods, two of which I can show you here. Graphing would probably be least useful in situations like these where large numbers with decimals are found. The other two methods, substitution and elimination, can be used effectively to isolate the variables and tell us what they equal. First, I will show you substitution in two ways:
1. Solving for variable s by manipulating variable a:
a + s = 321
a = 321 - s (subtract s to the other side to get a by itself)
$3.50(321-s) + $2.50s = $937.50 (substitution from step 2)
$1123.50 - $3.50s + $2.50s = $937.50 (distributive property)
$1123.50 - $1.50s = $937.50 (combine like terms)
-$1.00s = -$186 (subtraction property of equality)
s = 186 student tickets.
2. Solving for variable a by manipulating variable s:
a + s = 321
s = 321 - a
$3.50a + $2.50(321-a) = $937.50
$3.50a + $802.50 - $2.50 a = $937.50
$1.00a = $135
a = 135 adult tickets.
If you plug both numbers into the original equations, they should come out to be true equations. 135 + 186 = 321 and $472.5 + $465 = $937.5, so these two numbers are your answers for a and s accordingly.
Using the elimination method, we need to eliminate one of the variables by a process of subtraction or addition depending on the case. We can use multiplication or division before eliminating in order to eliminate.
a + s = 321
3.50a + 3.50s = 1123.50 (multiply entire equation by 3.50 in order to eliminate variable a by subtraction)
Now, we can use elimination:
3.50a + 3.50s = 1123.50
3.50a + 2.50s = 937.50
0a + 1s = 186, or s = 186 student tickets.
The same process can be used to solve for variable a. Your answers should be a = 135 adult tickets and s = 186 student tickets.
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