Math Homework? Word problems using either substitution or elimination, please answer!!!?

Question

Its only three problems, i had to ask for the other ones, and was shown how to do them, but the ones im trying now just wont work for me, :(

4) a group of 40 children attended a baseball game on a field trip. Each child received either a hot dog or a bag of popcorn. Hot dogs were $2.25 and popcorn was $1.75. If the total bill was $83.50, how many hot dogs and bags of popcorn were purchased.

the other two are in this picture. Number 7 and number 9, please help me on all of these, because i have tried and just can't seem to do it,

http://i159.photobucket.com/albums/t159/pieloverllamas/mathhomework001-1.jpg

please and thank you for any help you can give!!!

Answer 1

The answer to question 4 is 27 hotdogs and 13 popcorns.

Question 9 is 2 of the regular batteries and 3 packs of the alkaline.

Question 7 is 50 mylar and 75 latex balloons.

mrvadeboncoeur's got the right way to do it.  Kudos.

Answer 2

H = number of children who got hot dogs

P = number of children who got popcorn

(each child received either a hot dog or popcorn, none received both, none did not receive either)

total number of children = 40

total number of children = number who got hot dogs + number who got popcorn

H + P = 40

Price of a hot dog = $2.25

Price of popcorn = $1.75

Total bill = $83.50

price of hot dog * number of children who got hot dogs = $2.25*H

price of popcorn * number of children who got popcorn = $1.75*P

Total bill = $83.50 = $2.25*H + $1.75*P

so we have two equations:

eq. A:  H + P = 40

eq. B:  $83.50 = $2.25*H + $1.75*P

We have two equations, and two unknowns.  The number of equations is greater or equal to the number of unknowns, so we can solve this.

Remember that we can add/subtract/multiply/divide by the same amount to both sides of an equals equation, and still maintain the equality.  We want to solve for the unknown variable by getting the variable all by itself on one side of the equation and multiplied only by 1.

Substitution:

Let us re-write eq. A:

H + P = 40

H + P - P = 40 - P

H = 40 - P

now let us substitute this equation into eq. B:

$83.50 = $2.25*H + $1.75*P

$83.50 = $2.25*(40 - P) + $1.75*P

now we have one equation and one unknown, so we solve with basic algebra to find out the value for P:

$83.50 = $2.25*(40 - P) + $1.75*P

(100/$)*83.50 = (100/$)*[2.25*(40 - P) + 1.75*P]

8350 = 225*(40 - P) + 175*P

8350 = 9000 - 225*P + 175*P

8350 = 9000 - 50*P

8350 + 50*P = 9000 - 50*P + 50*P

8350 + 50*P = 9000

8350 + 50*P - 8350 = 9000 - 8350

50*P = 650

50*P/50 = 650/50

P = 13

now we plug in the value for P back into the derived equation from eq. A to find out the value for H:

H = 40 - P

H = 40 - 13

H = 27

So, there were 27 children that received hot dogs, and 13 children that received popcorn.

Check:

eq. A:

H + P = 40

27 + 13 ?= 40

40 == 40

eq. B:

$83.50 = $2.25*H + $1.75*P

$83.50 ?= $2.25*27 + $1.75*13

$83.50 ?= $60.75 + $22.75

$83.50 == $83.50

Question 7:

L + M = 125

$0.10*L + $0.50*M = $32.50

Question 9:

(kinda arbitrary example, since you shouldn't be mixing different battery types to power a device!  Might as well buy just the cheaper batteries, since you cannot buy 5 packages of the longer-lasting alkaline batteries for less than $25, and the stereo would just stop working when the cheaper mixed-in batteries stop working...)

R + A = 5

$4.25*A + $5.50*B = $25