Question

Marcy purchases a total of 12 adult, child, and senior tickets for an art museum tour. Adult tickets are $5 each, child tickets are $2 each, and senior tickets are $4 each. If she purchases 4 more child tickets than senior tickets and spends a total of $38, which statement is true?
The solution to this system is no viable because it results in fractional values of tickets.
The solution to this system is viable.
The solution to this system is nonviable because it results in negative values of tickets.
The solution to this system is no viable because it results in fractional values of money.

Answer 1

Answer:

A: The solution to this system is no viable because it results in fractional values of tickets

Step-by-step explanation:

Let number of adult, child, and senior tickets be x, y and z respectively.

Thus;

x + y + z = 12 - - - (eq 1)

she purchases 4 more child tickets than senior tickets.

Thus, y = z + 4

Thus:

x + (z + 4) + z = 12

x + 2z + 4 = 12

x + 2z = 8 - - - (eq 2)

Also, We are told that Adult tickets are $5 each, child tickets are $2 each, and senior tickets are $4 each. She spends a total of $38,

Thus;

5x + 2(z + 4) + 4z = 38

5x + 2z + 8 + 4z = 38

5x + 6z + 8 = 38

5x + 6z = 38 - 8

5x + 6z = 30

Divide through by 5 to get;

x + (6/5)z = 6 - - - (eq 3)

Subtract eq 2 from eq 1 to get;

2z - (6/5)z = 8 - 6

2z - (6/5)z = 2

Multiply through by 5 to get;

10z - 6z = 10

4z = 10

z = 2½

It's not possible to have a fractional value of a ticket and thus we can say that the solution is not viable.