document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=215117>Question 215117</A>: <I><SPAN STYLE=\"word-break: break-all;\"> One of two complementary angles is ywo more than one third of the other. Find the two angles. \r<BR>\n" );
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document.write( "I set up the equation as 90=x+(1/3x)+2 and proceeded to solve but i got x as equalling 132. This cant be right since complementary angles only can equal 90. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #162650 by </B><A HREF=/tutors/aboutme.mpl?userid=RAY100><B>RAY100(1637)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RAY100><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=162650\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Very  good  ,,,,just  off  by  a  negative  sign<BR>\n" );
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document.write( "Let  x &  y  be  complementary  angles,,,x+y=90,,,or,,,x=90-y<BR>\n" );
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document.write( "y= 2 + x/3,,,,,subst,,,,,y= 2+(90-y)/3 = 2+30-y/3<BR>\n" );
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document.write( "1.333 y = 32<BR>\n" );
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document.write( "y=24,,,,x=90-24 = 66<BR>\n" );
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document.write( "check\r<BR>\n" );
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document.write( "24 = 2 + 66/3,,,,ok<BR>\n" );
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