Asked by Anonymous
                solve the inequality: x-3/x-2 + 12/x^2-x-2> 8/X+1
            
            
        Answers
                    Answered by
            bobpursley
            
    multiply both sides by (x-2)(x+1)
(x-3)(x+1)+12>8
x^2-2x-3+4>0
(x-1)^2>0
which means that x can be any value except 1
    
(x-3)(x+1)+12>8
x^2-2x-3+4>0
(x-1)^2>0
which means that x can be any value except 1
                    Answered by
            Steve
            
    Hmmm. I get
(x-3)/(x-2) + 12/(x^2-x-2) > 8/(x+1)
(x-3)(x+1) + 12 > 8(x-2)
x^2-2x-3+12 > 8x-16
x^2-10x+25 > 0
(x-5)^2 > 0
So, all x except x = 2,5 and -1
Oddly enough, it appears that the real solution is x < -1 or x > 2 and x not 5. See
http://www.wolframalpha.com/input/?i=%28x-3%29%2F%28x-2%29+%2B+12%2F%28x^2-x-2%29+%3E+8%2F%28x%2B1%29+for+-2+%3Cx+%3C+6
Maybe you can spot my math error. Probably involves clearing the fractions, since the direction of the inequality will change with negative values.
    
(x-3)/(x-2) + 12/(x^2-x-2) > 8/(x+1)
(x-3)(x+1) + 12 > 8(x-2)
x^2-2x-3+12 > 8x-16
x^2-10x+25 > 0
(x-5)^2 > 0
So, all x except x = 2,5 and -1
Oddly enough, it appears that the real solution is x < -1 or x > 2 and x not 5. See
http://www.wolframalpha.com/input/?i=%28x-3%29%2F%28x-2%29+%2B+12%2F%28x^2-x-2%29+%3E+8%2F%28x%2B1%29+for+-2+%3Cx+%3C+6
Maybe you can spot my math error. Probably involves clearing the fractions, since the direction of the inequality will change with negative values.
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