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anonymous

  • 5 years ago

show work too: x+2/x-3 - 1/x = 3/x^2-3x fyi answer is -1

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  1. anonymous
    • 5 years ago
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    \[{x+2 \over x-3}-{1 \over x}={3 \over x(x-3)} \implies {x(x+2)-(x-3) \over x(x-3)}={3 \over x(x-3)}\] Both sides have the same denominator, so for for all values other than \(x=0\) and \( x=3\) \[x(x+2)-(x-3)=3 \implies x^2+2x-x+3=3 \implies x^2+x=0 \implies x(x+1)=0 \] So either \(x=0\) or \(x=-1\). But \(x=0\) is rejected, hence the only solution is \(x=-1\).

  2. gw2011
    • 5 years ago
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    (x+2)/(x-3) - 1/x = 3/(x^2-3x) (x+2)/(x-3) - 1/x = 3/(x)(x-3) (x+2)/(x-3) - 1/x - 3/(x)(x-3) = 0 Now find a common denominator [(x+2)(x) - (x-3) -3]/)x)(x-3) = 0 (x^2+2x-x+3-3)/(x)(x-3) = 0 (x^2+x)/(x)(x-3) = 0 (x)(x+1)/(x)(x-3) = 0 (x+1)/(x-3) = 0 Multiply both sides of the equation by (x-3) and you get: x+1 = 0 x = -1

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