The ice rink sold 90 tickets for the afternoon session, for a total of $700. General admission tickets cost
$10 and youth tickets cost $6 each. How many of each kind of tickets were sold?
College Algebra - Steve today at 6:45pm
g+y = 90
10g+6y = 900 I am confused.
Looks to me like Steve let the number of general admission tickets be g
and the number of youth tickets be y
How many total general and youth tickets were there??
g+y = 90
What was the cost of those tickets ??
10g + 6y = 900
What part did you not understand?
You have two equations in two unknowns, judging by the nature of the question you must know how to solve those.
Do you solve for y first then substitute y into 10g +6y =900. Help.
OOPS
looks like Steve misread the question and used 900 instead of $700, and I just copied his equation
so..... the second equation should have been
10g + 6y = 700
from g+y = 90
y = 90-g
now into the 2nd:
10g + 6y = 700
10g + 6(90-g) = 700
10g + 540 - 6g = 700
look, no more y, all g's
4g = 160
g = 40
Now back into y = 90-g = 90-40 = 50
so we had 40 general admission and 50 youths
Maybe oops. This question has been posted more than once, with differing numbers, just like the almonds/macadamia question and the walking/cycling questions.
To solve the system of equations for this problem, we need to use a method called substitution.
Let's start by rearranging the first equation, g + y = 90, to solve for g: g = 90 - y.
Next, we substitute this value of g into the second equation, 10g + 6y = 900, to eliminate g:
10(90 - y) + 6y = 900
900 - 10y + 6y = 900
-4y = 0
Simplifying further, we get -4y = 0. To solve for y, we divide by -4 on both sides of the equation:
y = 0.
Now that we have the value of y, we can substitute it back into the first equation to find g:
g + 0 = 90
g = 90.
Therefore, the solution is g = 90 and y = 0.
This means that no youth tickets were sold, and all 90 tickets sold were general admission tickets.