MHB Solve the inequality and graph the solution a real number line

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To solve the inequality 5/(x-1) - (2x)/(x+1) - 1 < 0, the expression can be manipulated by multiplying through by (x-1)²(x+1)² to maintain positivity. This leads to the simplified form -3x² + 7x + 6 < 0, which can be factored and analyzed using sign diagrams. The critical points from the factorization are x = -1, x = -2/3, and x = 3. The solution intervals are identified as x < -1, -2/3 < x < 1, and x > 3, providing a complete understanding of the inequality's solution on the real number line.
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5/(x-1) - (2x)/(x+1) - 1 < 0

How does one solve this inequality?
 
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megacat8921 said:
5/(x-1) - (2x)/(x+1) - 1 < 0

How does one solve this inequality?

you can multiply by $(x-1)^2(x+1)^2$ (kindly note squared to have it positive and get

$5(x-1)(x+1)^2 - 2x(x-1)^2(x+1) - (x-1)^2(x+1)^2 \lt 0$
expand and factor LHS to get the result
 
Getting a common denominator and simplifying gives $$\dfrac{-3x^2+7x+6}{(x-1)(x+1)}<0$$.

Now construct sign diagrams for $$-3x^2+7x+6$$ and $$x^2-1$$.

You should arrive at $$x<-1,-\dfrac23<x<1$$ and $$x>3$$.
 
kaliprasad said:
you can multiply by $(x-1)^2(x+1)^2$ (kindly note squared to have it positive and get

$5(x-1)(x+1)^2 - 2x(x-1)^2(x+1) - (x-1)^2(x+1)^2 \lt 0$
expand and factor LHS to get the result

Why expand and factor? You can already pick out common factors...

$\displaystyle \begin{align*} 5 \left( x - 1 \right) \left( x + 1 \right) ^2 - 2x \left( x - 1 \right) ^2 \left( x + 1 \right) - \left( x - 1 \right) ^2 \left( x + 1 \right) ^2 &= \left( x - 1 \right) \left( x + 1 \right) \left[ 5 \left( x + 1 \right) - 2x \left( x - 1 \right) - \left( x - 1 \right) \left( x + 1 \right) \right] \\ &= \left( x- 1 \right) \left( x + 1 \right) \left( 5x + 5 - 2x^2 + 2x - x^2 + 1 \right) \\ &= \left( x - 1 \right) \left( x + 1 \right) \left( - 3x^2 + 7x + 6 \right) \\ &= \left( x - 1 \right) \left( x + 1 \right) \left( -3x^2 + 9x - 2x + 6 \right) \\ &= \left( x - 1 \right) \left( x + 1 \right) \left[ -3x \left( x - 3 \right) - 2 \left( x - 3 \right) \right] \\ &= - \left( x - 1 \right) \left( x + 1 \right) \left( x - 3 \right) \left( 3x + 2 \right) \end{align*}$
 
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