document.write( "Question 372647: How would I solve this equation?\r
\n" ); document.write( "\n" ); document.write( " x - 1 = 2
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\n" ); document.write( "x+1 x-1 x^2-1\r
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\n" ); document.write( "\n" ); document.write( "I used the LCD of x+1, x-1 and i know the answer cannot be x=1 or x=-1. I came up with the answer of x= -1 but since it cannot be that, i put the empty set symbol of 0. Please help me, if this is incorrect? thank you.
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Algebra.Com's Answer #265465 by Jk22(389)\"\" \"About 
You can put this solution on YOUR website!
the LCD of x-1 and x+1 is (x+1)(x-1)=(x^2-1)
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\n" ); document.write( "Domain of definition x is not 1 or -1.
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\n" ); document.write( "\n" ); document.write( "\"+x%2F%28x%2B1%29-1%2F%28x-1%29=2%2F%28x%5E2-1%29\" times (x^2-1)=(x+1)(x-1)\r
\n" ); document.write( "\n" ); document.write( "x(x-1) - (x+1) = 2
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\n" ); document.write( "x^2-x-x-1 = 2
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\n" ); document.write( "x^2 -2x -3 = 0
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\n" ); document.write( "(x-3)(x+1)=0
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\n" ); document.write( "the solutions are : x=3 or x=-1, x=-1 is no solution since it would be a division by zero)
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\n" ); document.write( "Solution : x=3
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\n" ); document.write( "Verification : 3/4 - 1/2 = 1/4
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\n" ); document.write( "and RHS : 2/(9-1)=2/8=1/4, which is the same.
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