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.
I have seen a Question already answered Today which was this, and the Answer is B.
Answer:
D.
Step-by-step explanation:
I'm going to have to say C.
Answer:
30 dollars i think
Step-by-step explanation:
