document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=970613>Question 970613</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am getting confused to do this question please help me to solve this. Hoping for your cooperation.<BR>\n" );
document.write( "1. Solve:-<BR>\n" );
document.write( "a. y-1/y+1=2y-3/2y+3<BR>\n" );
document.write( "b. 2/y-1+3/y+1=5/y </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #593205 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=593205\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> first one:\r<BR>\n" );
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document.write( "(y-1)/(y+1) = (2y-3) / (2y+3)\r<BR>\n" );
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document.write( "cross multiply to get (y-1) * (2y+3) = (y+1) * (2y-3)\r<BR>\n" );
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document.write( "simplify to get 2y^2 + 3y - 2y - 3 = 2y^2 -3y + 2y - 3\r<BR>\n" );
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document.write( "combine like terms to get 2y^2 + y - 3 = 2y^2 - y - 3\r<BR>\n" );
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document.write( "subtract 2y^2 from both sides of the equation and add y to both sides of the equation and add 3 to both sides of the equation to get:\r<BR>\n" );
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document.write( "2y^2 - 2y^2 + y + y -3 + 3 = 2y^2 - 2y^2 - y + y - 3 + 3\r<BR>\n" );
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document.write( "combine like terms to get:\r<BR>\n" );
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document.write( "2y = 0 \r<BR>\n" );
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document.write( "divide both sides of the equation by 2 to get y = 0\r<BR>\n" );
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document.write( "that's your solution for the first one.\r<BR>\n" );
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document.write( "second one:\r<BR>\n" );
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document.write( "2/(y-1) + 3/(y+1) = (5/y)\r<BR>\n" );
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document.write( "multiply both sides of the equation by y*(y-1)*(y+1) to get:\r<BR>\n" );
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document.write( "2/(y-1)*y*(y-1)*(y+1) + 3/(y+1)*y*(y-1)*(y+1) = (5/y)*y*(y-1)*(y+1)\r<BR>\n" );
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document.write( "simplify to get:\r<BR>\n" );
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document.write( "2*y*(y+1) + 3*y*(y-1) = 5*(y-1)*(y+1)\r<BR>\n" );
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document.write( "simplify to get:\r<BR>\n" );
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document.write( "2y^2 + 2y + 3y^2 - 3y = 5y^2 - 5\r<BR>\n" );
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document.write( "combine like terms to get:\r<BR>\n" );
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document.write( "5y^2 - y = 5y^2 - 5\r<BR>\n" );
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document.write( "subtract 5y^2 from both sides of the equation to get:\r<BR>\n" );
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document.write( "5y^2 - 5y^2 - y = 5y^2 - 5y^2 - 5\r<BR>\n" );
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document.write( "combine like terms to get:\r<BR>\n" );
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document.write( "-y = -5\r<BR>\n" );
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document.write( "solve for y to get y = 5.\r<BR>\n" );
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document.write( "that's your solution for the second one.\r<BR>\n" );
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document.write( "solution for the first one is y = 0<BR>\n" );
document.write( "solution for the second one is y = 5\r<BR>\n" );
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document.write( "confirm by replacing y in the respective equations by the value for y in that eqution to see if the equations are true.\r<BR>\n" );
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document.write( "(y-1)/ (y+1) = (2y-3) / (2y+3) becomes:\r<BR>\n" );
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document.write( "-1 / 1 = -3 / 3 which becomes:\r<BR>\n" );
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document.write( "-1 = -1\r<BR>\n" );
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document.write( "first equation is true so the folution for the first problem is good.\r<BR>\n" );
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document.write( "2 / (y-1) + 3 / (y+1) = 5 / y becomes:\r<BR>\n" );
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document.write( "2 / 4 + 3 / 6 = 5 / 5\r<BR>\n" );
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document.write( "2/4 = 1/2 and 3/6 = 1/2 and 5/5 = 1, so you get:\r<BR>\n" );
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document.write( "1/2 + 1/2 = 1 which is true so the solution for the second problem is also good.\r<BR>\n" );
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document.write( "your solution to the first problem is y = 0.<BR>\n" );
document.write( "your solution to the second problem is y = 5.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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