document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=65569>Question 65569</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 4y           +        2y<BR>\n" );
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document.write( "y^2 + 6y + 5       y^2 - 1\r<BR>\n" );
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document.write( "I am just so lost on how to understand these types of problems! I become frustrated because I always end up with the wrong answer. Is there any hope in understanding this type of problem? Thank you for the great help :-) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #46189 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley71><B>checkley71(8403)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley71><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=46189\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 4y/(y+5)(y+1)+2y/(y+1)(y-1) multiply the second fraction by (y+5) & multiply ther first fraction by (y-1) thus you now have a common denominator & can now add the numerators. <BR>\n" );
document.write( "[(2y)(y+5)+4y(y-1)]/(y+5)(y+1)(y-1)<BR>\n" );
document.write( "(2y^2+5y+4y^2-4y)/(y-5)(y+1)(y-1)<BR>\n" );
document.write( "6y^2+y)/(y-5)(y+1)(y-1) factor out the y fron the numerator.<BR>\n" );
document.write( "y(6y+1)/(y+5)(y+1)(y-1)<BR>\n" );
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document.write( "MERRY CHRISTMAS </SPAN>\n" );
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