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.

Complete Question

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.

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      

substituting values

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

substituting values

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

given that

i.e  $56  $ 80

and  

i.e   $ 86 $ 100

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

1 and 2

step-by-step explanation: