On the first day of ticket sales, the school sold 3 adult tickets and 9 student tickets for a total of $75. The second day, the school sold $67 by selling 8 adu

Question

On the first day of ticket sales, the school sold 3 adult tickets and 9 student tickets for a total of $75. The second day, the school sold $67 by selling 8 adult and 5 student tickets. What is the price each of one adult and one student ticket?

Answer 1

A total of 586 people attended a banquet at Disney. A total of $9,423.70 in revenue was collected. Student tickets cost $12.95 and Adult tickets cost $17.95. Find the number of student and adult tickets sold.

Let: Student tickets = S

Let: Adult tickets = A

Given: Student ticket prices = $12.95

Given: Adult ticket prices = $17.95

Given: Total tickets sold = 586

Given: Total revenue = $9,423.70

S + A = 586                               (1)

12.95 S + 17.95 A = 9,423.70    (2)

Multiply equation (1) by –12.95

–12.95 S  – 12.95 A = –12.95 (586)

12.95 S + 17.95 A = 9,423.70

–12.95 S  – 12.95 A = –7,588.70

12.95 S + 17.95 A = 9,423.70

Add the two equations together. The S variables will drop out. That will leave us with A variables only. We then solve for A and substitute and solve for S.

5A = 1,835

Divide through by 5

A = 367 Adult tickets sold

Now substitute 367 for A and solve for S.

S + A = 586

S + 367 = 586

S = 219  Student tickets sold

Check 1

S + A = 586

219 + 367 = 586

586 = 586 Check

Check 2

12.95 S + 17.95 A = 9,423.70

12.95 (219) + 17.95 (367) = 9,423.70

2,836.05 + 6,587.65 = 9,423.70

9,423.70 = 9,423.70 Check

The final answer is, there were 219 Student tickets sold and 367 Adult tickets sold.

.

Answer 2

We have 2 unknowns, the cost of adult tickets and the cost of student tickets. Therefore, we need 2 equations to solve for this.

Let's let x = the cost of an adult ticket

            y = the cost of a student ticket

Eq. I    On the first day, the number of adult tickets times the cost of an adult ticket plus the number of student tickets                times the cost of a student tickets has to equal 75.

           3x + 9y = 75

Eq. II   Likewise, on the second day the number of adult tickets times the cost of an adult ticket plus the number                        of student tickets times the cost of a student ticket has to equal 67.

           8x + 5y =67

We can use either the substitution or elimination methods to solve this. Let's use substitution. In Eq. I we can simplify it by dividing everything by 3:

           3x + 9y = 75                     Eq. I

           (3x/3) + (9y/3) = 75/3       divide all by 3

           x + 3y = 25                       simplify

           x = 25 - 3y                        isolate x

Using this value for x we can substitute it in Eq. II and solve for y:

          8x + 5y = 67                      Eq. II

          8(25-3y) + 5y = 67             substitute for x

         200-24y + 5y =67                simplify

         200-67 = 24y-5y                  combine like terms

         133 = 19y                           simplify

         133/9 = y                           divide both sides to isolate y

           7 = y                                answer

Use this value for y in either equation and solve for x. Let's use Eq. I:

         3x + 9y = 75                       Eq. I

         3x +(9)(7) = 75                  substitute for y

         3x + 63 = 75                      simplify

         3x = 75 - 63                       isolate x term

         3x = 12                              simplify

           x = 4                                answer

We have x = 4 and y = 7. Let's check our answers by putting these values in for each equation:

Eq. I  3x + 9y = 75

         3(4) + 9(7) = 75

         12 + 63 = 75

         75 = 75

         Check

Eq. II  8x + 5y = 67

          8(4) + 5(7) = 67

          32 + 35 = 67

          67 = 67

          Check

The cost of an adult ticket is $4 and the cost of a student ticket is $7.