SOLUTION: A wholesaler is offering two different package deals of roses and carnations to florists. One package contains 20 dozen roses and 34 dozen carnations for $504.00. The other package

Question 1045376: A wholesaler is offering two different package deals of roses and carnations to florists. One package contains 20 dozen roses and 34 dozen carnations for $504.00. The other package contains 15 dozen roses and 17 dozen carnations for $327.00. Let x = cost of one dozen roses Let y = cost of one dozen carnations
a. Write a system of equations for this problem.
b. Which equation can be multiplied by a coefficient to create an opposite coefficient? What is that value?
c. Solve this system for the cost of one dozen roses and the cost of one dozen carnations.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
x = cost of one dozen roses.
y = cost of one dozen carnations.

20x + 34y = 504
15x + 17y = 327

you need to solve these 2 equations simultaneously.

if you multiply 17 by 2, you get 34.
therefore, the coefficient of y is a good candidate.

multiply both sides of the second equation by 2 and leave the first equation as is to get:

20x + 34y = 504
30x + 34y = 654

subtract the second equation from the first to get:

-10x = -150

solve for x to get x = 15.

replace x with 15 in either of the original equations and solve for y.

you will get y = 6

price of a dozen roses is 15 and price of a dozen carnations is 6.

note that i could also have done the following and gotten the same result.

start with:

20x + 34y = 504
15x + 17y = 327

multiply both sides of the second equation by -2 and leave the first equation as is to get:

20x + 34y = 504
-30x - 34y = -654

add the 2 equations together to get:

-10x = -150

solve for x to get x = 150.

if i multiplied by plus 2, then i subtracted one equation from the other to eliminate the y variable.

if i multiplied by minus 2, then i added one equation to the other to eliminate the y variable.

both methods work.

choose the one you feel more comfortable with.

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