SOLUTION: Valentine's Day will soon be here. Zarkada's Candy Store on Rhodes Avenue wants to create a mix of chocolates for $10.05per pound. It will use chocolate covered caramels that cost
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Question 406876: Valentine's Day will soon be here. Zarkada's Candy Store on Rhodes Avenue wants to create a mix of chocolates for $10.05per pound. It will use chocolate covered caramels that cost $10 per pound;turtles that cost $12 per pound; and chocolate kisses that cost $9 per pound. They want to use eight pounds more of the kisses than the turtles. If they decide to make 40 pounds of this candy mix, how much of each should they include.
This problem is Applications of systems of linear equations
solve the system of equations using any method
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Candy Store on Rhodes Avenue wants to create a mix of chocolates for $10.05 per pound.
It will use chocolate covered caramels that cost $10 per pound;
turtles that cost $12 per pound;
and chocolate kisses that cost $9 per pound.
They want to use eight pounds more of the kisses than the turtles.
If they decide to make 40 pounds of this candy mix, how much of each should they include.
:
Let c = amt of caramels
Let t = amt of turtles
Let k = amt of kisses
:
Write an equation for each statement:
:
"If they decide to make 40 pounds of this candy mix,"
c + t + k = 40
:
" use eight pounds more of the kisses than the turtles."
k = (t+8)
:
"create a mix of chocolates for $10.05 per pound."
10c + 12t + 9k = 10.05(40)
10c + 12t + 9k = 402
:
Multiply the 1st equation by 10 subtract from the above equaiton
10c + 12t + 9k = 402
10c + 10t + 10k= 400
-----------------------Subtraction eliminates c
0 + 2t - 1k = 2
Replace k with (t+8)
2t - (t+8) = 2
2t - t - 8 = 2
t = 2 + 8
t = 10 lb of turtles
then
k = t+8
k = 10+8
k = 18 lb kisses
:
Find the amt of caramels:
40 - 10 - 18 = 12 lb of Caramels
:
:
Check the solutions
10c + 12t + 9k = 10.05(40)
10(12) + 12(10) + 9(18) = 402
120 + 120 + 162 = 402, confirm our solution
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