The admission fee at an amusement park is $ 1.50 for children and $ 4.00 for adults. On a certain day, 258 people entered the park, and the admission fees collected totaled $712 . How many children and how many adults were admitted?

Question

I tried doing it this way;
x = amount of children tickets
y= amount of adults tickets

x + y = 258

1.50(x) + 4.00(y) = 712
1.50(x) + (y) = 712/4
1.50(x) + (y) = 178

1.50(x) + (y) = 178
- (x) + (y) = 258
-----------------------
0.50(x) + 0(y) = -80

x = -160

(x) + (y) = 258
-160 + (y) = 258

y= 258 + 160
y= 418

but that doesn't make any sense, and if I go
y= 258 - 160
y= 98
it show that its a wrong answer

Where did I make the mistake?

Answer 1

here is your problem:

1.50(x) + 4.00(y) = 712
1.50(x) + (y) = 712/4

you divided the 2 last terms by 4, but not the first one

Answer 2

so 1.50(x)/4 + 4.00(y)/4 = 712/4

0.375(x) + (y) = 178
(x) + (y) = 258
---------------------------------
-0.625(x) + 0(y) = -80

x= -80/-0.625
x= 128

x + y = 258
128 + y = 258
y = 258 - 128
y = 130

x = 128 (number of children tickets)
y = 130 (number of adults tickets)

And it works, Thank you Reiny!

Answer 3

The admission fee at an amusement park is $2.00 for children and $6.60 for adults. On a certain day, 299 people entered the park, and the admission fees collected totaled 1311 dollars. How many children and how many adults were admitted

Answer 4

The admission fee at an amusement park is $2.00 for children and $6.60 for adults. On a certain day, 299 people entered the park, and the admission fees collected totaled 1311 dollars. How many children and how many adults were admitted

Answer 5

A rectangular garden is 15 ft longer than it is wide. Its area is 1000ft^2
. What are its dimensions?