document.write( "Question 69181This question is from textbook Algebra Structure and Method
\n" ); document.write( ": Solve. If problem has no solution; write no solution. \r
\n" ); document.write( "\n" ); document.write( "1 = 3 over y+2 + 1 over y-2\r
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\n" ); document.write( "\n" ); document.write( "Here is what I have:\r
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\n" ); document.write( "\n" ); document.write( "(y+2)(y-2) = 3(y-2)+ y + 2\r
\n" ); document.write( "\n" ); document.write( "y^2 + 2y - 2y - 4 = 3y - 6 + y + 2\r
\n" ); document.write( "\n" ); document.write( "y^2 - 4 = 4y - 4\r
\n" ); document.write( "\n" ); document.write( "and that is as far as I got, I am lost
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Algebra.Com's Answer #49235 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
You did fine as far as you went so let's continue from where you left off:
\n" ); document.write( "\"y%5E2-4+=+4y-4\" Add 4 to both sides of the equation.
\n" ); document.write( "\"y%5E2+=+4y\" Subtract 4y from both sides.
\n" ); document.write( "\"y%5E2-4y+=+0\" Factor a y on the left side.
\n" ); document.write( "\"y%28y-4%29+=+0\" Apply the zero product principle.
\n" ); document.write( "\"y+=+0\" and/or \"y-4+=+0\"
\n" ); document.write( "The solutions are:
\n" ); document.write( "\"y+=+0\"
\n" ); document.write( "\"y+=+4\"
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