document.write( "Question 474472: Solve using the substitution method.\r
\n" ); document.write( "\n" ); document.write( "-x - 3y = -20
\n" ); document.write( "5x + 3y = 40 \n" ); document.write( "
Algebra.Com's Answer #325405 by mananth(16946)\"\" \"About 
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-1 x -3 y = -20 .............1
\n" ); document.write( "3 y = 40 -5 x ---------2
\n" ); document.write( "/ 3 =
\n" ); document.write( " y = 13.33 -1.67 x
\n" ); document.write( "Substitute y in (1)
\n" ); document.write( "-1x-3(13.33-1.67x)=-20
\n" ); document.write( "-1x-40+ 5x=-20
\n" ); document.write( "-1x+5x =-20+40
\n" ); document.write( "4x=20
\n" ); document.write( "/ 4
\n" ); document.write( "x= 5
\n" ); document.write( "Plug the value of x in (1)
\n" ); document.write( "-1 x -3 y = -20
\n" ); document.write( "-1 * 5 -3 y = -20
\n" ); document.write( "-5-3y=-20
\n" ); document.write( "-3y=-15
\n" ); document.write( "/-3
\n" ); document.write( "y=5
\n" ); document.write( " \n" ); document.write( "
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