document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1043899>Question 1043899</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Why do I have to use different steps for the same kind of problem?<BR>\n" );
document.write( "For x=a+b-2ab solve \"a\", all I have to do is flip it so it reads -2ab+b+a=x and then I can solve normally. <BR>\n" );
document.write( "But for x-a=a(y-b) solve \"a\", after I +a to both sides and end up with <BR>\n" );
document.write( "x=a(y-b)-a I can't flip it to -a+a(y-b)=x because that gives a different answer than using the additive property of equality. Why can't both be flipped when a is only on 1 side? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #659093 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=659093\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> When you add +a to both sides, you end up with x=a(y-b) PLUS a.<BR>\n" );
document.write( "You don't have to move it anywhere but factor out an a and divide by whatever was left behind after factoring.<BR>\n" );
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document.write( "x=a+b-2ab<BR>\n" );
document.write( "isolate the a<BR>\n" );
document.write( "x-b=a-2ab<BR>\n" );
document.write( "factor an a out of the right side<BR>\n" );
document.write( "x-b=a(1-2b)<BR>\n" );
document.write( "divide by (1-2b) both sides<BR>\n" );
document.write( "(x-b)/(1-2b)=a<BR>\n" );
document.write( "==================<BR>\n" );
document.write( "x-a=a(y-b)<BR>\n" );
document.write( "distribute<BR>\n" );
document.write( "x-a=ay-ab<BR>\n" );
document.write( "add a to both sides<BR>\n" );
document.write( "x=ay-ab+a<BR>\n" );
document.write( "factor out an a on the right<BR>\n" );
document.write( "x=a(y-b+1)<BR>\n" );
document.write( "divide both sides by (y+b+1)<BR>\n" );
document.write( "a=x/(y+b+1)<BR>\n" );
document.write( "=================<BR>\n" );
document.write( "The approach to the problem is to usually distribute everything and then take non (in this case) factors of a and move them to the other side.  Then factor an a out of everything that contains it, and divide both sides by what is left over.  In some ways, it is the same kind of problem, solving for a variable.  In other ways, it isn't, because the a is distributed over another pair of variables or constants, and that has to be dealt with.  You could in the second add the a and have x=a(y-b)+a.  That would be factored to x=a(y-b+1), because what was added happened to be a itself. </SPAN>\n" );
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