document.write( "Question 110799: Solve the system by addition or substitution.\r
\n" ); document.write( "\n" ); document.write( "3x + 6y = 0
\n" ); document.write( "x = 2/3
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Algebra.Com's Answer #80789 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "Start with the given system\r
\n" ); document.write( "\n" ); document.write( "\"3x%2B6y=0\"
\n" ); document.write( "\"x=2%2F3\"\r
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\n" ); document.write( "\n" ); document.write( "\"3%282%2F3%29%2B6y=0\" Plug in \"x=2%2F3\" into the first equation. In other words, replace each \"x\" with \"2\". Notice we've eliminated the \"x\" variables. So we now have a simple equation with one unknown.\r
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\n" ); document.write( "\n" ); document.write( "\"2%2B6y=0\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "\"6y=0-2\"Subtract 6 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"6y=-2\" Combine like terms on the right side\r
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\n" ); document.write( "\n" ); document.write( "\"y=%28-2%29%2F%286%29\" Divide both sides by 6 to isolate y\r
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\n" ); document.write( "\n" ); document.write( "\"y=-1%2F3\" Reduce\r
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\n" ); document.write( "\n" ); document.write( "So our answer is \"x=2%2F3\", \"y=-1%2F3\"
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