document.write( "Question 198299This question is from textbook Intermediate Algebra An Applied Approach
\n" ); document.write( ": 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. \n" ); document.write( "
Algebra.Com's Answer #148764 by ankor@dixie-net.com(22740)\"\" \"About 
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A merchant mixed 10 lb of cinnamon tea with 5 lb of spice tea.
\n" ); document.write( " The 15 pd mixture cost $40.
\n" ); document.write( ":
\n" ); document.write( "A second mixture included 12 lb of the cinnamon tea and 8 lb of the spice tea.
\n" ); document.write( ":
\n" ); document.write( " The 20 pd mixture cost $54. Find the cost per pound of the cinnamon tea and of the spice tea.
\n" ); document.write( ":
\n" ); document.write( "Let c = cost/lb for cinnamon tea
\n" ); document.write( "Let s = cost/lb for spice tea
\n" ); document.write( ":
\n" ); document.write( "Two equations
\n" ); document.write( "10c + 5s = 40
\n" ); document.write( "12c + 8s = 54
\n" ); document.write( ":
\n" ); document.write( "Multiply 1st equation by 8, the 2nd equation by 5, results:
\n" ); document.write( "80c + 40s = 320
\n" ); document.write( "60c + 40s = 270
\n" ); document.write( "-------------------subtraction eliminates s, find c
\n" ); document.write( "20c = 50
\n" ); document.write( "c = \"50%2F20\"
\n" ); document.write( "c = $2.50 a lb for cinnamon tea
\n" ); document.write( ":
\n" ); document.write( "Find s, substitute 2.5 for c in the 1st equation
\n" ); document.write( "25 + 5s = 40
\n" ); document.write( "5s = 40 - 25
\n" ); document.write( "s = \"15%2F5\"
\n" ); document.write( "s = $3 for spice tea
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\n" ); document.write( "Check solution 2nd equation
\n" ); document.write( "12(2.5) + 8(3) =
\n" ); document.write( "30 + 24 = 54
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