Answer:
First city was= $ x = $ 5604
Second city was =$(x +1000) = $6604
Explanation:
The tax in the first city was 8.5%
The tax in the second city was 5.5%
Let the hotel charge amount before tax for the first city be $x
Let the hotel charge amount before tax for the second city be: $(x+1000)
Applying the tax
First city will be;
8.5/100*(x) = $0.085 x
Second city will be;
5.5/100* (x+100) =0.055(x+100) = $ (0.055x+5.5)
Total hotel tax paid for the two cities was ;
0.085 x + 0.055 x + 5.5 = $(0.14 x + 5.5)
This expression for total hotel tax paid for the two cities is equal to $790
$(0.14 x + 5.5) = $ 790
0.14 x = 790 -5.5
0.14 x = 784.5
x= 784.5/0.14
x=5603.57
x= $5604 ( nearest $)
Before tax, hotel charge for;
First city was= $ x = $ 5604
Second city was =$(x +1000) = $6604
The group paid $ 5250 at first city and $ 6250 at second city
Solution:Let x = the charge in 1st city before taxesLet y = the charge in 2nd city before taxesThe hotel charge before tax in the second city was $1000 higher than in the firstThen the charge at the second hotel before tax will be x + 1000y = x + 1000 ----- eqn 1The 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 $790Therefore, a equation is framed as:8.5 % of x + 5.5 % of y = 7900.085x + 0.055y = 790 ------- eqn 2Let us solve eqn 1 and eqn 2Substitute eqn 1 in eqn 20.085x + 0.055(x + 1000) = 7900.085x + 0.055x + 55 = 7900.14x = 790 - 550.14x = 735x = 5250Substitute x = 5250 in eqn 1y = 5250 + 1000y = 6250Thus the group paid $ 5250 at first city and $ 6250 at second city
