The Eskimos are trying to generate more money because player wages are increasing. they have came up with a couple of plans:
Plan 1 : they are going to open up 1850 new seats in the endzone that will be reserved for family seating. they project that they will, on average, get 1850 adults and children to fill those seats if they charge $4.50 per child and $7.50 per adult and they should make $10125.00
a)how many adults and children are needed per game (solve using substitution)
b)how many money do they generate over the course of a 16 game season?
Plan 2: they are planning to sell 75 year anniversary logo football to help raise money. an original full size logo football will sell for $22 and a miniature logo football will sell for $6. if they sell all 6000 footballs they ordered they can make $76000.
c) how many of each type of football did they order?(solve using elimination)
My Work: is it right?
ok a = adult, c = children
10125.00/4.50 = 4.50+7.50a
-2250 children and 1350 adults 3600 adults and children are needed per game
b)10125(16) = $162.000
c)76000 = 22f+6f
3455+2545
Work is incorrect. See details below.
a) a+c=1850 and $7.50a+$4.50c=$10,125. Substitute a=1850-c which gives
7.50(1850-c)+4.50c=10,125
13,875-7.50c+4.50c=10,125
3.00c=3,750
c=1,250
a=1,850-1,250=600
b) 10,125x16=$162,000
c) F=full size football
M=miniature football
F+M=6,000
$22F+$6M=$76,000
Substitute for M=6,000-F
22F+6(6,000-F)=76,000
22F+36,000-6F=76,000
16F=40,000
F=2,500
M=6,000-F=6,000-2,500=3,500
Thank you, your the best :)
a) To solve using substitution, let's set up the equations:
Let a be the number of adults and c be the number of children.
From the information given, we know that:
1850 = c + a (equation 1)
$4.50c + $7.50a = $10,125 (equation 2)
To solve using substitution, we can solve equation 1 for c and substitute it into equation 2.
From equation 1, we have:
c = 1850 - a
Substituting c into equation 2, we get:
$4.50(1850 - a) + $7.50a = $10,125
Expanding and simplifying, we have:
$8,325 - $4.50a + $7.50a = $10,125
Combine like terms:
$3.00a = $1,800
Divide both sides by $3:
a = 600
Substituting the value of a back into equation 1:
1850 = c + 600
c = 1250
Therefore, they need 600 adults and 1250 children per game.
b) To calculate the money generated over the course of a 16-game season, we multiply the money generated per game by the number of games in a season:
$10,125 x 16 = $162,000
So, they will generate $162,000 over the course of a 16-game season.
c) To solve using elimination, let's set up the equations:
Let f be the number of original full-size logo footballs and m be the number of miniature logo footballs.
From the information given, we know that:
f + m = 600 (equation 1)
$22f + $6m = $76,000 (equation 2)
To solve using elimination, let's multiply equation 1 by 6 to match the coefficients of m in both equations:
6f + 6m = 3600
Now we can eliminate m by subtracting equation 1 from this new equation:
(6f + 6m) - (f + m) = 3600 - 600
5f + 5m = 3000
Combining like terms, we have:
5f + 5m = 3000
22f + 6m = 76,000
If we multiply equation 1 by 22, we can eliminate f:
22f + 22m = 13,200
Now we can eliminate f by subtracting equation 1 from this new equation:
(22f + 22m) - (22f + 6m) = 13,200 - 76,000
16m = 62,800
Divide both sides by 16:
m = 3,925
Substituting the value of m back into equation 1:
f + 3,925 = 600
f = 600 - 3,925
f = -3,325
We cannot have a negative number of footballs, so this solution is not valid.
Therefore, there is no solution using the elimination method for this problem.