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
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-> 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
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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):
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.