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2 answers:
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The group paid $ 5250 at first city and $ 6250 at second city
<u>Solution:</u>
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
<em><u>Therefore, a equation is framed as:</u></em>
8.5 % of x + 5.5 % of y = 790
0.085x + 0.055y = 790 ------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
<em><u>Substitute eqn 1 in eqn 2</u></em>
0.085x + 0.055(x + 1000) = 790
0.085x + 0.055x + 55 = 790
0.14x = 790 - 55
0.14x = 735
<h3>x = 5250</h3><em><u>Substitute x = 5250 in eqn 1</u></em>
y = 5250 + 1000
<h3>y = 6250</h3>Thus the group paid $ 5250 at first city and $ 6250 at second city