document.write( "Question 588367: i need to solve for this equation:
\n" ); document.write( "((x+1)/x-2))+((x-3)/(x-1))<0
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Algebra.Com's Answer #374495 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
\"%28x%2B1%29%2F%28x-2%29%2B%28x-3%29%2F%28x-1%29%3C0\"
\n" ); document.write( "Maybe we could get a common denominator:
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\n" ); document.write( "And we could do those multiplications in the numerator:
\n" ); document.write( "\"%28x%5E2-1%29%2F%28%28x-2%29%28x-1%29%29%2B%28x%5E2-5x%2B6%29%2F%28%28x-2%29%28x-1%29%29%3C0\"
\n" ); document.write( "And we could add the two fractions:
\n" ); document.write( "\"%282x%5E2-5x%2B5%29%2F%28%28x-2%29%28x-1%29%29%3C0\"
\n" ); document.write( "The numerator is always positive.
\n" ); document.write( "We can see that \"2x%5E2-5x%2B5=0\" has no real roots because the discriminant is negative:
\n" ); document.write( "\"%28-5%29%5E2-4%2A2%2A5=25-40=-15%3C0\"
\n" ); document.write( "\"2x%5E2-5x%2B5\" is positive for x=0 (it's 5), and everywhere else.
\n" ); document.write( "For \"%282x%5E2-5x%2B5%29%2F%28%28x-2%29%28x-1%29%29\" to be negative, the denominator has to be negative.
\n" ); document.write( "Each factor has a zero (x=1 and x=2), is negative for lesser x values and positive for greater values.
\n" ); document.write( "The product will be negative for x such that \"1%3Cx%3C2\".
\n" ); document.write( "For x=1 and x=2, the denominator is zero and the functions do not exist.
\n" ); document.write( "For x<1 and x>2 the function is positive.
\n" ); document.write( "So the solution is \"highlight%281%3Cx%3C2%29\" \n" ); document.write( "
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