document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=832783>Question 832783</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need you to solve these systems by substitution\r<BR>\n" );
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document.write( "Y=0.5x+2<BR>\n" );
document.write( "-y=-2x+4 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #502185 by </B><A HREF=/tutors/aboutme.mpl?userid=LinnW><B>LinnW(1048)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=LinnW><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=LinnW><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=502185\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> y=0.5x+2<BR>\n" );
document.write( "-y=-2x+4\r<BR>\n" );
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document.write( "substitute 0.5x+2 for y in -y=-2x+4<BR>\n" );
document.write( "-(0.5x+2)=-2x+4<BR>\n" );
document.write( "-0.5x - 1 = -2x + 4<BR>\n" );
document.write( "add 2x to each side<BR>\n" );
document.write( "1.5x -1 = 4<BR>\n" );
document.write( "add 1 to each side<BR>\n" );
document.write( "1.5x = 3<BR>\n" );
document.write( "divide each side by 1.5<BR>\n" );
document.write( "x = 2\r<BR>\n" );
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document.write( "Since y = 0.5x + 2 , y = 1 + 2 = 3<BR>\n" );
document.write( "y = 3 </SPAN>\n" );
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