document.write( "Question 253952: x+y-2z=0
\n" ); document.write( "3x+y=1
\n" ); document.write( "5x+3y+7z=2\r
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Algebra.Com's Answer #186326 by dabanfield(803)\"\" \"About 
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1)x+y-2z=0
\n" ); document.write( "2)3x+y=1
\n" ); document.write( "3)5x+3y+7z=2 \r
\n" ); document.write( "\n" ); document.write( "From equation 1 we have y=2z-x \r
\n" ); document.write( "\n" ); document.write( "So if (from equation 2) 3x+y = 1 then\r
\n" ); document.write( "\n" ); document.write( "3x+(2z-x)= 1\r
\n" ); document.write( "\n" ); document.write( "3x+2z-x = 1 \r
\n" ); document.write( "\n" ); document.write( "and \r
\n" ); document.write( "\n" ); document.write( "From equation 3:\r
\n" ); document.write( "\n" ); document.write( "5x+3*(2z-x) + 7z = 2\r
\n" ); document.write( "\n" ); document.write( "Solve these two equations for x an z then substitute in the first equation to find y.\r
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