document.write( "Question 643911: x+y=9.0
\n" ); document.write( "0.50x+.20y=3.75\r
\n" ); document.write( "\n" ); document.write( "With the given given problem, solve for the value of x and y.\r
\n" ); document.write( "\n" ); document.write( "The answer I continue to get is x=4.3 and y=4.7 but it is wrong. Help please?
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Algebra.Com's Answer #404694 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
1.\"x%2By=9.0\"
\n" ); document.write( "2.\"0.50x%2B.20y=3.75\"
\n" ); document.write( "From eq. 1,
\n" ); document.write( "\"y=9-x\"
\n" ); document.write( "Use this in eq. 2,
\n" ); document.write( "\"0.50x%2B0.2%289-x%29=3.75\"
\n" ); document.write( "\"0.50x%2B1.8-0.2x=3.75\"
\n" ); document.write( "\"0.30x%2B1.8=3.75\"
\n" ); document.write( "\"0.30x=+1.95\"
\n" ); document.write( "\"x=1.95%2F0.30\"
\n" ); document.write( "\"highlight%28x=6.5%29\"
\n" ); document.write( "Now use the substitution again,
\n" ); document.write( "\"y=9-x\"
\n" ); document.write( "\"y=9-6.5\"
\n" ); document.write( "\"highlight%28y=2.5%29\"
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