document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=572140>Question 572140</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 14/(x+4)=5-x </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #368664 by </B><A HREF=/tutors/aboutme.mpl?userid=mathsmiles><B>mathsmiles(68)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mathsmiles><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=mathsmiles><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=368664\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 14/(x+4)=5-x\r<BR>\n" );
document.write( "\n" );
document.write( "The first thing I'd do is get rid of the (x+4) so we don't have a fraction anymore<BR>\n" );
document.write( "Do that by multiplying both sides by (x+4) giving ... \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "14 = (5-x)(x+4)  Now multiply out the parens using FOIL ...<BR>\n" );
document.write( "14 = 5x + 20 - x^2 - 4x  Rearranging the terms ... <BR>\n" );
document.write( "14 = -x^2 + 5x - 4x + 20  Now combine the x terms ... <BR>\n" );
document.write( "14 = -x^2 + x + 20   Now let's get all the factors on one side of the equation so we can solve.  Subtract 14 from both sides ...<BR>\n" );
document.write( "-x^2 + x + 6 = 0  I like to make the x^2 term positive, so let's multiply by (-1)<BR>\n" );
document.write( "(-1)(-x^2 + x + 6) = 0<BR>\n" );
document.write( "x^2 - x - 6 = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Now we need to factor into two multipliers:<BR>\n" );
document.write( "(x _ _)(x _ _) = 0  where each _ represents something we need to figure out, both the operator and the term\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Since the last term is negative, we know we need one of each + and - operators:<BR>\n" );
document.write( "(x - _)(x + _) = 0\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Now we need factors of 6 which are different by 1 to get the -x terms:<BR>\n" );
document.write( "Factors of 6:<BR>\n" );
document.write( "6 = 2 * 3  That'll do it!\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Since we want the middle x term to be negative, let's place 3 in the negative paren giving:<BR>\n" );
document.write( "(x - 3)(x + 2) = 0<BR>\n" );
document.write( "Now we have two possibilities for x:<BR>\n" );
document.write( "x-3 = 0  Adding 3 to each side:<BR>\n" );
document.write( "x = 3<BR>\n" );
document.write( "OR<BR>\n" );
document.write( "x+2 = 0  Adding -2 to each side:<BR>\n" );
document.write( "x = -2\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Since your problem didn't specify positive or negative, either of these should work.  Let's check:<BR>\n" );
document.write( "14/(x+4)=5-x  Substituting x=3<BR>\n" );
document.write( "14/(3+4)= 5-3<BR>\n" );
document.write( "14/7 = 5-3<BR>\n" );
document.write( "2 = 2 Correct!\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "14/(x+4) = 5-x  Substituting x=-2:<BR>\n" );
document.write( "14/(-2+4) = 5-(-2) Converting the double negative on the right to positive:<BR>\n" );
document.write( "14/(2)=5+2<BR>\n" );
document.write( "14/2 = 7<BR>\n" );
document.write( "7=7  Correct!\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );