document.write( "Question 562122: I need an equation for the tollowing word problem.
\n" ); document.write( "at your lemonade stand you charge .50 for 1/2 cup and .75 for a full cup at the end of the day you see that 12 cups have been used and you have made 8$ how name of each size of drink did you sell? \n" ); document.write( "

Algebra.Com's Answer #364450 by mananth(16946) $\"\"$ $\"About$

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1/2 cup ---x numbers
\n" ); document.write( "full cup---y numbers\r
\n" ); document.write( "\n" ); document.write( "x+y =12\r
\n" ); document.write( "\n" ); document.write( "0.5x+0.75y=8
\n" ); document.write( "multiply by 100
\n" ); document.write( "50x+75y=800\r
\n" ); document.write( "\n" ); document.write( "these are the two equations solve for x & y \r
\n" ); document.write( "\n" ); document.write( "1 x + 1 y = 12 .............1
\n" ); document.write( "50 x + 75 y = 800 .............2
\n" ); document.write( "Eliminate y
\n" ); document.write( "multiply (1)by -75
\n" ); document.write( "Multiply (2) by 1
\n" ); document.write( "-75 x -75 y = -900
\n" ); document.write( "50 x 75 y = 800
\n" ); document.write( "Add the two equations
\n" ); document.write( "-25 x = -100
\n" ); document.write( "/ -25
\n" ); document.write( "x = 4 1/2 cup
\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "1 x + 1 y = 12
\n" ); document.write( "4 + 1 y = 12
\n" ); document.write( " 1 y = 12 -4
\n" ); document.write( " 1 y = 8
\n" ); document.write( " y = 8 full cup \n" ); document.write( "

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