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:
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 : [tex]x+y=553[/tex] ---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 : [tex]6x+9y=3936[/tex] ---B
Plot A and B
[tex]x+y=553[/tex] ---Green
[tex]6x+9y=3936[/tex] ---Purple
Intersection point = (x,y)=(347,206)
Refer the attached figure
Hence 347 child tickets and 206 adult tickets are sold
