SOLUTION: The verbiage in this question is confusing me so much that I don't even know where to start. Any help in explaining and solving would be greatly appreciated. The following tabl

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Question 1177662: The verbiage in this question is confusing me so much that I don't even know where to start. Any help in explaining and solving would be greatly appreciated.
The following table shows the amount spent by four U.S. airlines to fly one available seat 1 mile in the second quarter of 2014. Set up a system and then solve using technology.
// The table shows:
Airline ==> United Continental - American - Jet Blue - Southwest
Cost(in cents)==> 14.9, 14.6, 11.9, 12.4

Suppose that, on a 3,000-mile New York to Los Angeles flight, United Continental, American, and Southwest flew a total of 235 empty seats, costing them a total of $99,060. If United Continental had three times as many empty seats as American, how many empty seats did each of these three airlines carry on its flight?
Found 3 solutions by CPhill, ikleyn, Edwin McCravy:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
You're right, the wording makes it a bit tricky! Let's break it down and solve it step-by-step.
**1. Define Variables**
* Let 'u' be the number of empty seats on United Continental.
* Let 'a' be the number of empty seats on American.
* Let 's' be the number of empty seats on Southwest.
**2. Convert Costs to Dollars**
* United Continental: 14.9 cents = $0.149 per seat-mile
* American: 14.6 cents = $0.146 per seat-mile
* Southwest: 12.4 cents = $0.124 per seat-mile
**3. Set Up the System of Equations**
* **Total Empty Seats:** u + a + s = 235
* **Total Cost of Empty Seats:** (0.149 * 3000)u + (0.146 * 3000)a + (0.124 * 3000)s = 99060
* **United Continental vs. American Empty Seats:** u = 3a
**4. Simplify the Cost Equation**
* 447u + 438a + 372s = 99060
**5. Substitute u = 3a into the Other Equations**
* 3a + a + s = 235 => 4a + s = 235
* 447(3a) + 438a + 372s = 99060 => 1341a + 438a + 372s = 99060 => 1779a + 372s = 99060
**6. Solve for 's' in the First Simplified Equation**
* s = 235 - 4a
**7. Substitute 's' into the Second Simplified Equation**
* 1779a + 372(235 - 4a) = 99060
* 1779a + 87420 - 1488a = 99060
* 291a = 11640
* a = 11640 / 291
* a = 40
**8. Solve for 'u' and 's'**
* u = 3a = 3 * 40 = 120
* s = 235 - 4a = 235 - 4 * 40 = 235 - 160 = 75
**Answer**
* United Continental: 120 empty seats
* American: 40 empty seats
* Southwest: 75 empty seats
Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.

        As the problem is solved in the post by @CPhill, it is one of possible ways/methods,
        but there is another, much more effective and more impressive way to solve the problem
        using only one equation from the very beginning, instead of three.

        Also, it is worth to notice one very important fact:
        in the problem the information about  4  companies is given,  but actually only three
        companies participate in the game.  The fourth company,  Jet Blue,  is introduced
        and does present only to confuse the reader and to distract attention. 


Let x be the number of empty seats for American.

Then the number of empty seats for United Continental is 3x,
and the number of empty seats for Southwest is (235 - x - 3x) = 235 - 4x.


Now write the total cost equation for empty seats

    United Cont.        American        Southwest
    3000*0.149*(3x)  +  3000*0.146*x  + 3000*0.124*(235-4x) = 99060  dollars.


Simplify and find x

     1341x            + 438x          + 87420 - 1488x  = 99060

                        291x                           = 99060 - 87420 = 11640.

                           x                           = 11640/191 = 40.


So, the number of empty seats for American is 40;  for United Continental  3*40 = 120,
                          and for Southwest 235-4*40 = 75.

Solved.

Notice that in this solution the system of three equations is not even mentioned.



Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
Ikleyn, This problem is numbered 1177662.  That means it was posted more 
than 4 years ago.  All problems numbered 1180000 or less are more than 4 years old.

How do I know?  I clicked on My Algebra.com (all my stuff) and found that
I answered problem 1180995 back in March, 2021.  That was 4 years ago, so
this problem, numbered 1177662 is older than that.

You can find all the solutions you've answered by clicking on 

My Algebra.com (all my stuff) 

at the top of THIS PAGE.

You can also find the date any problem you've answered was posted by clicking up
there. 

Edwin
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