Respuesta :
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
[tex]\frac{8.5}{100} \times x + \frac{5.5}{100} \times y = 790[/tex]
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
x = 5250
Substitute x = 5250 in eqn 1
y = 5250 + 1000
y = 6250
Thus the group paid $ 5250 at first city and $ 6250 at second city
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: brainly.com/question/23589883
