SOLUTION: Problem Page A theater group made appearances in two cities. The hotel charge before tax in the second city was $1000 higher than in the first. The tax in the first city was 5% , a

Question 1110786: Problem Page A theater group made appearances in two cities. The hotel charge before tax in the second city was $1000 higher than in the first. The tax in the first city was 5% , and the tax in the second city was 6% . The total hotel tax paid for the two cities was $692.50 . How much was the hotel charge in each city before tax?
Answer by ikleyn(52943) (Show Source): You can put this solution on YOUR website!
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Let x = The hotel charge before tax in the first city;

    y = the hotel charge before tax in the second city.


Then the problem says

   -x +     y = 1000,       (1)
0.05x + 0.06y =  692.50.    (2)


I will show you how to solve it by the Elimination method.

For it, multiply eq(1) by 0.05 (both sides). You will get


-0.05x + 0.05y = 50,        (1')
 0.05x + 0.06y = 692.50.    (2')


Now add equations ('1) and (2').  The terms  "-0.05x"  and "0.05x" will cancel each other, and you will get a single equation for only one unknown y

0.05y + 0.06y = 50 + 692.50,    or

0.11y = 742.50.  ================>  y =  = 6750.


Thus you found that in the second city the hotel charged 6750 dollars before tax.


Hence, in the first city the hotel charged 5750 dollars before tax.

Solved. // On the way, you learned on how the Elimination method works.

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