document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=514766>Question 514766</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a triangle is 72. The first side is 10 more than the second side.the third side is half of the first triangle. Find the length of each side. Please help I've done this so many times! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #343557 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=343557\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x, y and z be the lengths of the three sides (first, second, and third respectively)\r<BR>\n" );
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document.write( "x+y+z = 72\r<BR>\n" );
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document.write( "x+y+(1/2)x = 72\r<BR>\n" );
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document.write( "(y+10)+y+(1/2)(y+10) = 72\r<BR>\n" );
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document.write( "y+10+y+(1/2)y+5 = 72\r<BR>\n" );
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document.write( "(5/2)y+15=72\r<BR>\n" );
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document.write( "(5/2)y=72-15\r<BR>\n" );
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document.write( "(5/2)y=57\r<BR>\n" );
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document.write( "5y=57*2\r<BR>\n" );
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document.write( "5y=114\r<BR>\n" );
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document.write( "y=114/5\r<BR>\n" );
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document.write( "Now use this value of y to solve the equations\r<BR>\n" );
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document.write( "x = y+10\r<BR>\n" );
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document.write( "z = (1/2)x\r<BR>\n" );
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document.write( "to find the lengths of the other 2 sides </SPAN>\n" );
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