Can somone help me please?

You purchase 20 dozen roses and 15 dozen carnations at a cost of $504 for your school student council to use as a Valentine fundraiser. You package them together into bouquets of 2 roses for $5.00 or 3 carnations for $3.00. Altogether your student council sold 251 bouquets for a total of $989.00.





Let x = number of rose bouquets sold



Let y = number of carnation bouquets sold





Which of the following systems models the above situation?



a. x + y = 251 and 20x + 15y = 504

b. x + y = 251 and 5x + 3y = 989

c. x + y = 989 and 5x + 3y = 251

d. x + y = 251 and 20x + 15y = 989

Can somone help me please?
d
Reply:Lots of irrelevant info

but problem only gives two variables x %26amp; y bouq. sold

also gives total of x + y = 251

also says what the price for each is 5 %26amp; 3 for x%26amp; y resp.

If I bought 3 rose b's and 4 carn. b's how much would it cost me?

3 * 5 + 4 * 3 = 15 + 12 = $27 right?

so, for any amount of x %26amp; y I purchase the formula is

5x + 3y = cost

for the total cost = total sold = 5x+3y = 989

....

you can also go further and figure out how many of each was sold

if x+y = 251 then y = 251 - x

substituting that into the cost equation

5x + 3(251-x) = 989 or 5x + 753 - 3x = 989 = 2x +753

or 989 - 753 =2x = 236 so x = 118

if x = 118 then y = 251-118 = 133

also we can figure the profit (assuming no other costs)

989 - 504 = $485

but we CAN'T figure the cost to you for a rose (or a dozen) or carnations, we just know that 20*12*r + 15*12*c = 504

where r and c are your cost for buying the flowers. We could have r=2 %26amp; c = 2/15 or r=1 %26amp; c = 22/15 or r=1.5 %26amp; c=.75 or an infinite number of other possibilities. If we knew that roses cost twice as much as carnations then we could solve it (see above). Just need another clue.
Reply:a
Reply:NONE!