Question

A theater group made appearances in two cities. The hotel charge before tax in the second city was $500 higher than in the first. The tax in the first city was 8.5%, and the tax in the second city was 7.5%. The total hotel tax paid for the two cities was $917.50. How much was the hotel charge in each city before tax?First city:Second city:Solve by tax rate or interest rate using system of linear equations.

Answer 1

Answer:

• First city: $5,500

,

• Second city: $6,000

Explanation:

• Let the hotel charge before tax in the first city = x

,

• Let the hotel charge before tax in the second city = y

The hotel charge before tax in the second city was $500 higher than in the first.

$y=x+500\cdots(1)$

The tax in the first city was 8.5%, and the tax in the second city was 7.5%.

The total hotel tax paid for the two cities was $917.50.

$\begin{gathered} (8.5\%\text{ of x\rparen+\lparen7.5\% of y\rparen=917.50} \\ 0.085x+0.075y=917.50\cdots(2) \end{gathered}$

The problem gives rise to the system of linear equations below:

$\begin{gathered} y=x+500\cdots(1) \\ 0.085x+0.075y=917.50\operatorname{\cdots}(2) \end{gathered}$

The system is then solved for x and y.

Substitute equation (1) into equation (2):

$\begin{gathered} 0.085x+0.075y=917.50\operatorname{\cdots}(2) \\ 0.085x+0.075(x+500)=917.50 \\ 0.085x+0.075x+37.50=917.50 \\ 0.16x=917.50-37.50 \\ 0.16x=880 \\ \text{ Divide both sides by 0.16} \\ \frac{0.16x}{0.16}=\frac{880}{0.16} \\ x=\ 5500 \end{gathered}$

Finally, solve for y:

$\begin{gathered} y=x+500 \\ =5500+500 \\ =\ 6,000 \end{gathered}$

The hotel charge before tax in each city was:

• First city: $5,500

,

• Second city: $6,000