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