Question 11.1
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a theater group made appearances in two cities. the hotel charge before tax is the
second city was $1000 higher than in the first. the tax in the first city was 8.5% and the
tax in the second city was 5.5%. the total hotel tax paid for the two cities was $790.
how much was the hotel charge in each city before tax?

The hotel charge in the first city  is $24,500 and the  hotel charge in the second city is.

$25,500Two equations can be derived from this question:

x - y = 1000 equation 1

0.055x + 0.085y = 790 equation 2

Where:

y = hotel charge in the first city

x = hotel charge in the second city

In order to determine the value of y, take the following steps:

Multiply equation 1 by 0.055

0.055x - 0.055y = 55 equation 3

Subtract equation 3 from 2

735 = 0.030y

Divide both sides by 0.030

y = $24,500

Substitute for y in equation 1

x -  $24,500 = 1000

x = $25,500

To learn more about simultaneous equations, please check: link

The group paid $ 5250 at first city and $ 6250 at second city

Solution:

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

The hotel charge before tax in the  second city was $1000 higher than in the first

Then the charge at the second hotel before tax will be x + 1000

y = x + 1000 ----- eqn 1

The tax in the first city was 8.5% and the  tax in the second city was 5.5%

The total hotel tax paid for the two cities was $790

Therefore, a equation is framed as:

8.5 % of x + 5.5 % of y = 790

0.085x + 0.055y = 790 ------- eqn 2

Let us solve eqn 1 and eqn 2

Substitute eqn 1 in eqn 2

0.085x + 0.055(x + 1000) = 790

0.085x + 0.055x + 55 = 790

0.14x = 790 - 55

0.14x = 735

Substitute x = 5250 in eqn 1

y = 5250 + 1000

Thus the group paid $ 5250 at first city and $ 6250 at second city

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