document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=83114>Question 83114</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 10-(x-2)=15 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #59652 by </B><A HREF=/tutors/aboutme.mpl?userid=adamchapman><B>adamchapman(301)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=adamchapman><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=59652\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 10-(x-2)=15\r<BR>\n" );
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document.write( "subtracting \"what is in ther brackets\" is the same as \"adding everthing in the brackets multiplied by -1\"\r<BR>\n" );
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document.write( "10+(-x--2)=15\r<BR>\n" );
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document.write( "the double negative in front of the 2 makes a positive\r<BR>\n" );
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document.write( "10+(-x+2)=15\r<BR>\n" );
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document.write( "now we can get rid of the brakets\r<BR>\n" );
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document.write( "10-x+2=15\r<BR>\n" );
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document.write( "now  we need to do is add/subrtact until we get only x on one side of the equals sign:\r<BR>\n" );
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document.write( "10-x=13<BR>\n" );
document.write( "-x=3\r<BR>\n" );
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document.write( "so x=-3\r<BR>\n" );
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document.write( "I hope this helps, \r<BR>\n" );
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document.write( "Adam </SPAN>\n" );
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