Question

A theater group made appearances in two cities. The hotel charge before tax in the second city was $1500 higher than in the first. The tax in the first city was 4% , and the tax in the second city was 3.5% . The total hotel tax paid for the two cities was $408.75. How much was the hotel charge in each city before tax?

Answer 1

Answer:

• $4750 and $6250

Step-by-step explanation:

Given

• Hotel charge in city 1 = x
• Hotel charge in city 2 = y

The equation for the hotel charge:

• y = x + 1500

Taxes

• Tax in the city 1 = 4% = 0.04x
• Tax in the city 2 = 3.5% = 0.035y
• Total tax = $408.75

The equation for the tax:

• 0.04x + 0.035y = 408.75

Substitute y and solve for x:

• 0.04x + 0.035(x + 1500) = 408.75
• 0.04x + 0.035x + 52.5 = 408.75
• 0.075x = 408.75 - 52.5
• 0.075x = 356.25
• x = 356.25/0.075
• x = 4750

Find the value of y:

• y = 4750 + 1500
• y = 6250

The hotel charge was:

• The first city: $4750
• The second city: $6250

Answer 2

Sorry for the delay!

Let x = charge before tax in 1st city, and y = charge before tax in 2nd city

y = $1500 + x

4/100x => tax in first city

3.5/100y => tax in second city

0.04x + 0.035y = $408.75

We have the following system of equations to solve:

$\begin{bmatrix}y=1500+x\\ 0.04x+0.035y=408.75\end{bmatrix}\\\\\mathrm{Substitute\:}y=1500+x\\\\begin{bmatrix}0.04x+0.035\left(1500+x\right)=408.75\end{bmatrix}\\\begin{bmatrix}0.075x+52.5=408.75\end{bmatrix}\\\\\ x = 4750\\\mathrm{For\:}y=1500+x\\\mathrm{Substitute\:}x=4750:\\\\y=1500+4750 = 6250$

So the hotel charge before tax in city 1 = $4750, and the hotel charge before tax in city 2 = $6250