document.write( "Question 1178773: in the tea shop they prepare a tea mixture with a price of 165 euros per kilogram. they will use cherry-flavored tea for EUR 250 per kilogram and apple tea 150 per kilogram to prepare this mixture. how many grams of tea they need to prepare one kilogram of mixture. \n" ); document.write( "
Algebra.Com's Answer #808169 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
x+y=1\r
\n" ); document.write( "\n" ); document.write( "250x+150y=165(x+y)
\n" ); document.write( "rearrange\r
\n" ); document.write( "\n" ); document.write( "15x-85y=0\r
\n" ); document.write( "\n" ); document.write( "let x grams cherry flavoured tea
\n" ); document.write( " y grams of apple tea
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "1.00 x + 1.00 y = 1.00
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "250.00 x + 150.00 y = 165.00 .............2
\n" ); document.write( "Eliminate y
\n" ); document.write( "multiply (1)by -150.00
\n" ); document.write( "Multiply (2) by 1.00
\n" ); document.write( "-150.00 x -150.00 y = -150.00
\n" ); document.write( "250.00 x 150.00 y = 165.00
\n" ); document.write( "Add the two equations
\n" ); document.write( "100.00 x = 15.00
\n" ); document.write( "/ 100.00
\n" ); document.write( "x = 0.15
\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "1.00 x + 1.00 y = 1.00
\n" ); document.write( "0.15 + 1.00 y = 1.00
\n" ); document.write( " 1.00 y = 0.85
\n" ); document.write( " y = 0.85
\n" ); document.write( "Ans x = 0.15
\n" ); document.write( " y = 0.85
\n" ); document.write( " 150 grams cherry flavoured tea
\n" ); document.write( " 850 grams of apple tea \r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );