document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=114574>Question 114574</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the length of the second side of a trianglei s 2 inches less than the lenght of the first side. the length of the third side is 12 inches more than the length of the first side. the perimeter is 73 inches. find the lenth of each side of the triangle. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #83362 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=83362\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> PERIMETER OF A TRIANGLE=SIDE1+SIDE2+SIDE3<BR>\n" );
document.write( "LET X=THE FIRST SIDE.<BR>\n" );
document.write( "THE SECOND SIDE=X-2<BR>\n" );
document.write( "THE THIRD SIDE=X+12 <BR>\n" );
document.write( "73=X+X-2+X+12<BR>\n" );
document.write( "73=3X+10<BR>\n" );
document.write( "3X=73-10<BR>\n" );
document.write( "3X=63<BR>\n" );
document.write( "X=63/3<BR>\n" );
document.write( "X=21 FOR THE FIRST SIDE.<BR>\n" );
document.write( "21-2=19 FOR THE SECOND.<BR>\n" );
document.write( "21+12=33 FOR THE THIRD.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( "21+19+33=73<BR>\n" );
document.write( "73=73<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );