document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=175924>Question 175924</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve each system by addition.\r<BR>\n" );
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document.write( "2x=2-y<BR>\n" );
document.write( "3x+y=-1 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #131036 by </B><A HREF=/tutors/aboutme.mpl?userid=EMStelley><B>EMStelley(208)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=EMStelley><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=EMStelley><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=131036\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The first step here is to get both equations in the same form.  So let's make the first one look like the second one.  This can be accomplished by adding y to both sides to obtain<BR>\n" );
document.write( "2x+y=2<BR>\n" );
document.write( "So now we have to solve<BR>\n" );
document.write( "2x+y=2<BR>\n" );
document.write( "3x+y=-1<BR>\n" );
document.write( "In order to use the addition method, our goal must be to get one of the variables to \"go away\" by adding the two equations together.  Right now, that would not occur.  But notice that if we multiplied the second equation by -1, then the y's would cancel.  <BR>\n" );
document.write( "2x + y = 2<BR>\n" );
document.write( "3x - y = -1<BR>\n" );
document.write( "Adding, we get<BR>\n" );
document.write( "5x = 1<BR>\n" );
document.write( "So x = 1/5.  All that is left to do is to find y.  We can do this by substituting x = 1/5 into either equation.  I will use the first one<BR>\n" );
document.write( "2(1/5)+y=2<BR>\n" );
document.write( "2/5 + y = 2<BR>\n" );
document.write( "y = 2 - 2/5 = 8/5<BR>\n" );
document.write( "So the solution is (1/5,8/5) </SPAN>\n" );
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