Grandpa and Grandma are treating their family to the movies. Matinee tickets cost $4 per child and $4 per adult. Evening tickets cost $6 per child and $8 per adult. They plan on spending no more than $80 on the matinee tickets and no more than $100 on the evening tickets.

Grandpa and Grandma are treating their family to the movies. Matinee tickets cost $4 per child and $4 per adult. Evening tickets cost $6 per child and $8 per adult. They plan on spending no more than $80 on the matinee tickets and no more than $100 on the evening tickets. Could they take 9 children and 4 adults to both shows? Show your work. A yes or no answer is not sufficient for credit.

Answer:

Yes it is possible to take the 9 children and 4 adults to both shows

Step-by-step explanation:

From the question we are told that

The cost of the Matinee tickets for a child is z = $4

The cost of the Matinee tickets for an adult is a = $ 4

The cost of the Evening tickets for a child is k = $6

The cost of the Evening tickets for an adult is b = $8

The maximum amount to be spent on Matinee tickets is m = $80

The maximum amount to be spent on Evening tickets is e = $100

The number of child to be taken to the movies is n = 9

The number of adults to be taken to the movies is j = 4

Now the total amount of money that would be spent on Matinee tickets is mathematically evaluated as

[tex]t = 4 n + 4 j[/tex]

substituting values

[tex]t = 4 * 9 + 4* 4[/tex]

[tex]t = 52[/tex]

Now the total amount of money that would be spent on Evening ticket is mathematically evaluated as

[tex]T = 6n + 8j[/tex]

substituting values

[tex]T = 6(9) + 8(4)[/tex]

[tex]T = 86[/tex]

This implies that it is possible to take 9 children and 4 adults to both shows

given that

[tex]t \le m[/tex]

i.e $56 [tex]\le[/tex]$ 80

and

[tex]T \le e[/tex]

i.e $ 86 [tex]\le[/tex] $ 100