SOLUTION: A theater group made appearances in two cities. The hotel charge before tax in the second city was $500 lower than in the first. The tax in the first city was 8%, and the tax in th

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A theater group made appearances in two cities. The hotel charge before tax in the second city was $500 lower than in the first. The tax in the first city was 8%, and the tax in th      Log On


   

Question 1142571: A theater group made appearances in two cities. The hotel charge before tax in the second city was $500 lower than in the first. The tax in the first city was 8%, and the tax in the second city was 4%. The total hotel tax paid for the two cities was $640. How much was the hotel charge in each city before tax?
Answer by ikleyn(52903) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x = the hotel charge before tax in the first city;

    y = that of the second city.


Then from the condition you have these two equations in two unknowns


       x  -     y = 500     (1)

    0.08x + 0.04y = 640.    (2)


To solve the system of equations, use the Elimination method.


For it, from equation (1) express  x = 500 + y  and substitute it into the second equation. You will get


    0.08*(500+y) + 0.04y = 640.


Simplify and solve for y


    40 + 0.08y + 0.04y = 640

    0.12y = 640 - 40 = 600

    y = 600/0.12 = 5000.


ANSWER.  In the second city hotel charged $5000;  in the second city  $500 dollars more, or  5500  dollars.


CHECK.   0.08*5500 + 0.04*5000 = 640  dollars.    ! Correct !