CAN SOMEONE HELP ME?? I HAVE BEEN ASKING FOR THE PAST 3 DAYS

Question

A movie theater charges $7 for adults, $5 for children under age 17 and $4 for seniors over age 60. For one showing of The Jungle Book, the theater sold 222 tickets and took in $1383. If twice as many adults tickets were sold as the total of children and senior tickets, how many tickets from each type were sold?

Answer 1

Let x = number of  adult tickets

      y = number of children tickets

      z = number of senior tickets

 

Translate:

  “the theater sold 222 tickets”     means     x + y + z = 222     [eq1]

 

 “took in $1383”    means

      money rom  +  money from  +  money from     =    $1383  

      adult tickets   children tickets    senior tickets

          7x           +      5y                +       4z               =  1383   [eq2]

 

{note: For each class of ticket:    money = (cost per ticket)*(number of tickets)]

 

  “twice as many adults tickets were sold as the total of children and senior tickets”  means

      x = 2(y+z)    [eq3]

 

Now, the math:

      2x + 2y + 2z = 444      [this is eq1*2]

      2x + (2y + 2z) = 444     [now, look at eq3 as x = 2y + 2z]

         2x  + x  = 444

             3x = 444

             x = 148

 

There were 148 adult tickets sold.

 

Now, eq1 becomes:

   148 + y + z = 222

   y + z = 74                 [eq4]

 

And, eq2 becomes:

    1036  +  5y  + 4z  = 1383

 

     5y   +  4z   =  347

     4y   +  4z   =  296      [this is 4*eq4]

----------------------------    [elimination; subtract equations]

      y               =   51

There were 51 children’s tickets sold.

 

And now, eq1 is:

   148 + 51 + z = 222

       z = 23

There were 23 senior’s tickets sold.

 

Answer 2

This is a system of equations.  It has three variables, so there will be more work involved, but you can still solve it.  Let x = number of adult tickets, y = number of child tickets, and z = number of senior tickets.  Remember, when you purchase something, you multiply the price by the number sold and then add up the subtotals to get the total cost.  In this case, 7*x is the money for the adults, 5*y is the money from the children and 4*z is the money from the seniors and the grand total is $1383.  This gives us our first equation:

 

(1) 7x + 5y + 4z = 1383

 

The number of tickets sold is just the sum of the number of tickets sold to each group.  This provides us with the equation:

 

(2) x + y + z = 222

 

Finally, we know that twice as many adult tickets were sold as the sum of the other two groups.  This provides an equation:

 

x = 2(y + z)

 

which simplifies to:

 

(3) x = 2y + 2z

 

You can use equation three to substitute 2y + 2z in for x in the first two equations.

 

(4) 7(2y + 2z) + 5y + 4z =1383

(5) (2y + 2z) + y + z = 222

 

If you simplify the new equations, you get the following two variable system:

 

(6) 19y + 18z = 1383

(7) 3y + 3z = 222

 

This system will solve nicely through the elimination method, where you try to align the variables so the cancel.  If you multiply equation 7 by -6, you'd have 18z and -18z with cancel.  Equation 7 becomes:

 

(8) -18y - 18z = -1332

 

Combining equations 6 and 8 so that the z's cancel:

 

y = 51 (51 children tickets were sold.

 

You can then use your result and substitute back in for y in equation 6.  This allow you to solve for z:

 

19(51) + 18z = 1383

969 + 18z = 1383

18z = 414

z = 23  (23 senior tickets were sold).

 

Using equation 3, we can substitute our values for y and z to get a value for x.

 

x = 2(51) + 2(23) =148  (148 adult tickets were sold)