document.write( "Question 289444: I hate word problems, but here it goes.\r
\n" ); document.write( "\n" ); document.write( "The owner of a tea shop wants to mix three kinds of tea to make 100 oz of a mixture that will sell for $0.83 per oz. He uses Orange Pekoe, which sells for $0.80 per oz, Irish Breakfast, for $0.85 per oz, and Earl Grey, for $0.95 per oz. If he wants to use twice as much Orange Pekoe as Irish Breakfast, how much of each kind of tea should he use?\r
\n" ); document.write( "\n" ); document.write( "So far these are the equations I have come up with:
\n" ); document.write( "x + y + z = 100\r
\n" ); document.write( "\n" ); document.write( "0.80x + 0.85y + 0.95z = 0.83\r
\n" ); document.write( "\n" ); document.write( "x = 2y\r
\n" ); document.write( "\n" ); document.write( "I plugged in x = 2y into the first equation, like this:
\n" ); document.write( "2y + y + z = 100\r
\n" ); document.write( "\n" ); document.write( "But I still need to get rid of the z. So I add 2y + 1y and solve for z, which gives me z = -3y + 100 and then I plug it into the first equation again, so that I get this: 2y + y + -3y + 100 = 100\r
\n" ); document.write( "\n" ); document.write( "But it cancels out. What am I doing wrong? How do I solve this problem? \n" ); document.write( "
Algebra.Com's Answer #209719 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
The owner of a tea shop wants to mix three kinds of tea to make 100 oz of a mixture that will sell for $0.83 per oz. He uses Orange Pekoe, which sells for $0.80 per oz, Irish Breakfast, for $0.85 per oz, and Earl Grey, for $0.95 per oz. If he wants to use twice as much Orange Pekoe as Irish Breakfast, how much of each kind of tea should he use?
\n" ); document.write( "So far these are the equations I have come up with:
\n" ); document.write( "x + y + z = 100
\n" ); document.write( ":
\n" ); document.write( "This equation should be
\n" ); document.write( "0.80x + 0.85y + 0.95z = 0.83(100)
\n" ); document.write( "x = 2y
\n" ); document.write( "I plugged in x = 2y into the first equation, like this:
\n" ); document.write( "2y + y + z = 100
\n" ); document.write( "3y + z = 100
\n" ); document.write( "then
\n" ); document.write( "z = (100-3y)
\n" ); document.write( ":
\n" ); document.write( "Substitute for x and z in the cost equation, find y
\n" ); document.write( "0.80x + 0.85y + 0.95z = 0.83(100)
\n" ); document.write( ".8(2y) + .85y + .05(100-3y) = 83
\n" ); document.write( "1.6y + .85y + 5 - .15y = 83
\n" ); document.write( "1.6y + .85y - .05y = 83 - 5
\n" ); document.write( "2.6y = 78
\n" ); document.write( "y = \"78%2F2.6\"
\n" ); document.write( "y = 30 oz of Irish tea
\n" ); document.write( ":
\n" ); document.write( "I'm sure you can find x and z now,
\n" ); document.write( ":
\n" ); document.write( "Note, remember you cannot substitute in the same equation that you derived
\n" ); document.write( "substitution expression from. If you do, that's what happens.
\n" ); document.write( " Always substitute in the \"other\" equation.
\n" ); document.write( ":
\n" ); document.write( "Check your 3 solutions in the two original equations.
\n" ); document.write( " \n" ); document.write( "
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