document.write( "Question 109890: solve the system by additio or substitution
\n" ); document.write( "3x+6y=0
\n" ); document.write( "x= 2/3\r
\n" ); document.write( "\n" ); document.write( "I have no idea how to work this problem with a fraction. Please help.
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Algebra.Com's Answer #80083 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
3x+6y=0
\n" ); document.write( "x= 2/3
\n" ); document.write( ":
\n" ); document.write( "Just substitute 2/3 for x in the 1st equation, and find y
\n" ); document.write( ":
\n" ); document.write( "3(\"2%2F3\") + 6y =0
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\n" ); document.write( "The 3's cancel and you have;
\n" ); document.write( "2 + 6y = 0
\n" ); document.write( "6y = -2
\n" ); document.write( "y = \"-2%2F6\"
\n" ); document.write( "y = \"-1%2F3\"
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\n" ); document.write( "Check solution in the original equation:
\n" ); document.write( "3(\"2%2F3\") + 6(\"-1%2F3\") = 0
\n" ); document.write( " 2 - 2 = 0
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