System
- 10s +25t = 11700
- s -2t = 0
Solution
- s = 520
- t = 260
Let s and t represent single-entry and three-day tickets, respectively. These variables represent the numbers we're asked to find: "how many of each [ticket type] he sold."
We are given the revenue from each ticket type, and the total revenue, so we can write an equation based on the relation between prices, numbers sold, and revenue:
... 10s +25t = 11700 . . . . . equation for total revenue
We are also given a relation between the two number of tickets sold:
... s = 2t . . . . . . . . . . . . . . . twice as many single tickets were sold as 3-day
We can rearrange this second equation to put it into standard form. That makes it easier to see what to do to eliminate a variable.
... s -2t = 0 . . . . . . . . . . . . subtract 2t to put into standard form
So, our system of equations is ...
- 10s +25t = 11700
- s -2t = 0
<em>What </em>elimination<em> is all about</em>
The idea with "elimination" is to find a multiple of one (or both) equations such that the coefficients of one of the variables are opposite. Then, the result of adding those multiples will be to eliminate that variable.
Here, we can multiply the second equation by -10 to make the coefficient of s be -10, the opposite of its value in the first equation. (We could also multiply the first equation by -0.1 to achieve the same result. This would result in a non-integer value for the coefficient of t, but the solution process would still work.)
Alternatively, we can multiply the first equation by 2 and the second equation by 25 to give two equations with 50t and -50t as the t-variable terms. These would cancel when added, so would eliminate the t variable. (It seems like more work to do that, so we'll choose the first option.)
<em>Solution by elimination</em>
... 10s +25t = 11700 . . . . our first equation
... -10s +20t = 0 . . . . . . . second equation of our system, multiplied by -10
... 45t = 11700 . . . . . . . . .the sum of these two equations (s-term eliminated)
... t = 11700/45 = 260 . . . . . divide by the coefficient of t
... s = 2t = 520 . . . . . . . . . . use the relationship with s to find s
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<em>Solution using your number sense</em>
As soon as you see there is a relation between single-day tickets and 3-day tickets, you can realize that all you need to do is bundle the tickets according to that relation, then find the number of bundles. Here, 2 single-day tickets and 1 three-day ticket will bundle to give a package worth 2×$10 + $25 = $45. Then the revenue of $11700 will be $11700/$45 = 260 packages of tickets. That amounts to 260 three-day tickets and 520 single-day tickets.
(You may notice that our elimination solution effectively computes this same result, where "t" and the number of "packages" is the same value (since there is 1 "t" in the package).)