Question

a ticket office sold 553 tickets one day. the receipts totaled $3936. how many $9 adult tickets and how many $6 child tickets were sold​

Answer 1

This is the assumption method ;)

Take all tickets to be adult tickets

$9 x 553 = $4977

Then..... $4977 - $3936 = $1041

The difference between the child and adult ticket is $3, so divide $1041 by $3

$1041 divided by $3 = 347

347 is the number of child tickets sold

347 x $6 = $2082

So 553 - 347 = 206 ( no. of adult tickets )

206 x $9 = $1854

Child tickets sold : 347

Adult tickets sold : 206

:0 ITS MAGIC IM SO SMART

Answer 2

Answer:

347 child tickets and 206 adult tickets are sold

Step-by-step explanation:

Let x be the no. of child tickets sold

Let y be the no. of adult tickets sold

A ticket office sold 553 tickets one day.

Equation becomes : $x+y=553$ ---A

Cost of 1 child ticket = $6

Cost of x child tickets = 6x

Cost of 1 adult ticket= $9

Cost of y adult tickets = 9y

Now we are given that the receipts totaled $3936.

So, Equation becomes : $6x+9y=3936$ ---B

Plot A and B

$x+y=553$ ---Green

$6x+9y=3936$ ---Purple

Intersection point = (x,y)=(347,206)

Refer the attached figure

Hence 347 child tickets and 206 adult tickets are sold