Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Can someone help me with this plz? Plz show ur work .... A local high school collected $1590 from 321 people who attended a football game. The price of each adult admission is $6. People between the ages of 4-17 paid a children admission rate of $4. How many adult tickets and child tickets were sold that day?

Sagot :

To solve this problem we can use system of equations.
First equation can be
321=x+y 
where x - ppl between 4-17 and y - adult
second equation can be as follows
1590=6*y+4*x
So we have
[tex] \left \{ {{321=x+y} \atop {1590=6y+4x}} \right. [/tex]
From first equation we get value of x in times of y
x=321-y
Now we can substitute it into second eq.
1590=6y+4(321-y) 
now simply if
1590=6y+1284-4y            /-1284 both sides
306=2y                /:2 divide both sides by 2
y=153 
Now we can back substitude
x=321-153
x=168
So we get result
[tex] \left \{ {{y=153 - adult} \atop {x=168 - kids}} \right. [/tex]

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.