document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=713727>Question 713727</A>: <I><SPAN STYLE=\"word-break: break-all;\"> One of two complementary angles is 6 degrees smaller than twice the other angle. Find the measure of each angle.\r<BR>\n" );
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document.write( "I have tried:\r<BR>\n" );
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document.write( "2x-6=180 -> add 6 to both sides<BR>\n" );
document.write( "2x=186 -> divide both sides by 2<BR>\n" );
document.write( "x = 93\r<BR>\n" );
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document.write( "And:\r<BR>\n" );
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document.write( "90-2x-6=180 -> add like terms<BR>\n" );
document.write( "84-2x=180 -> subtract 84 from both sides<BR>\n" );
document.write( "-2x=106 -> divide both sides by -2<BR>\n" );
document.write( "x=93\r<BR>\n" );
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document.write( "According to the back of my book the answers should be 32 and 58 but I can't seem to get there and it doesn't show how to do it.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #438675 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=438675\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the two angles be A and B.<BR>\n" );
document.write( "A = 2B - 6 [6 degrees smaller than twice the other angle]<BR>\n" );
document.write( "Since the two angles are complementary, A + B = 90 -> A = 90 - B<BR>\n" );
document.write( "Solve for B:<BR>\n" );
document.write( "90 - B = 2B - 6<BR>\n" );
document.write( "3B = 96<BR>\n" );
document.write( "B = 32<BR>\n" );
document.write( "Therefore A = 90 - 32 = 58<BR>\n" );
document.write( "Ans: 32, 58 </SPAN>\n" );
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