I need help please? >.<?

Question

C and Y candy company mixes candy that costs $6 per kg with candy that costs $4.50 per kg. how many kilograms of each are needed to make a 3 kg box that cost $15?

I did two other problems my method, and i did da same right here in the pic: http://i156.photobucket.com/albums/t10/tammygurl94/Untitledurmama.png
But it doesn't make sense... Help please?

Oh dont mind the bottom... I asked a friend for help as well... she didn't get it

Answer 1

Let X be the number of kilograms of $6 candy and Y be the number of kilograms of $4.50 candy.

X + Y = 3
X ( 6.00 ) + Y ( 4.50 ) = 15.00

So X = 3 - Y and

( 3 - Y ) ( 6.00 ) + Y ( 4.50 ) = 15.00
18.00 - 6.00 Y + 4.50 Y = 15.00
3.00 = 1.50 Y
2.00 = Y

So you use 2 kg of the cheap stuff and 1 kg of the more expensive candy to get a 3-kg box costing $15.00.

Answer 2

OK
I am assuming you mean a 3kg box that cost $15 for the box.

6x + 4.5y = 15
x+y = 3 ; let's use substitution

y= 3-x

6x + 4.5(3-x) = 15
6x +13.5 -4.5x = 15
6x -4.5x = 1.5
1.5x = 1.5
x = 1

So if x =1 ; y = 3-(1) ; y = 2

Check

6(1) + 4.5(2) = 15??
6 + 9 = 15??
15 = 15 YES!!

Hope this helps.

Answer 3

sorry, I don't use that method.
candy @ $6/kg = [(15 -- 3*4.5) / (6 -- 4.5)] kg = 1 kg
candy @ $4.5/kg = (3 -- 1) kg = 2kg