document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=447031>Question 447031</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me solve this problem on tangents and secants: \r<BR>\n" );
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document.write( "x^2 + y^2 = 10<BR>\n" );
document.write( "y = 3x </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #327835 by </B><A HREF=/tutors/aboutme.mpl?userid=cleomenius><B>cleomenius(959)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=cleomenius><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=cleomenius><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=327835\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Since y = 3x, we can use a substitution in the first equation.\r<BR>\n" );
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document.write( "x^2 +{3x)^2  = 10\r<BR>\n" );
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document.write( "10x^2   = 10<BR>\n" );
document.write( "x^2 = 1<BR>\n" );
document.write( "x = 1\r<BR>\n" );
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document.write( "Substituting into the first equation, we get\r<BR>\n" );
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document.write( "1 + y^2  = 10<BR>\n" );
document.write( "y^2  = 9<BR>\n" );
document.write( "y = 3.\r<BR>\n" );
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document.write( "This checks with the second equation, 3 = (3)(1).\r<BR>\n" );
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document.write( "Cleomenius.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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