SOLUTION: A merchant mixed 10 lb of cinnamon tea with 5 lb of spice tea. The 15 pd mixture cost $40. A second mixture included 12 lb of the cinnamon tea and 8 lb of the spice tea. The 20 pd

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Question 198299This question is from textbook Intermediate Algebra An Applied Approach
: A merchant mixed 10 lb of cinnamon tea with 5 lb of spice tea. The 15 pd mixture cost $40. A second mixture included 12 lb of the cinnamon tea and 8 lb of the spice tea. The 20 pd mixture cost $54. Find the cost per pound of the cinnamon tea and of the spice tea. This question is from textbook Intermediate Algebra An Applied Approach

Found 2 solutions by arallie, ankor@dixie-net.com:
Answer by arallie(162) About Me  (Show Source):
You can put this solution on YOUR website!
Make an algebra problem
10c%2B5t=40
12c%2B8t=54
Solve for a variable and substitute
10c%2B5t=40--->t=8-2c
12c%2B8%288-2c%29=54
12c%2B64-16c=54
10=4c
c=2.5
Substitute
12%282.5%29%2B8t=54
30%2B8t=54
8t=24
t=3
Check
10%282.5%29%2B5%283%29=40
25%2B15=40
40=40
cost per pound of the cinnamon tea $2.50
and of the spice tea $3.00

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A merchant mixed 10 lb of cinnamon tea with 5 lb of spice tea.
The 15 pd mixture cost $40.
:
A second mixture included 12 lb of the cinnamon tea and 8 lb of the spice tea.
:
The 20 pd mixture cost $54. Find the cost per pound of the cinnamon tea and of the spice tea.
:
Let c = cost/lb for cinnamon tea
Let s = cost/lb for spice tea
:
Two equations
10c + 5s = 40
12c + 8s = 54
:
Multiply 1st equation by 8, the 2nd equation by 5, results:
80c + 40s = 320
60c + 40s = 270
-------------------subtraction eliminates s, find c
20c = 50
c = 50%2F20
c = $2.50 a lb for cinnamon tea
:
Find s, substitute 2.5 for c in the 1st equation
25 + 5s = 40
5s = 40 - 25
s = 15%2F5
s = $3 for spice tea
;
;
Check solution 2nd equation
12(2.5) + 8(3) =
30 + 24 = 54