Solve the inequality and graph the solution a real number line

[megacat8921]

5/(x-1) - (2x)/(x+1) - 1 < 0

How does one solve this inequality?

 

[kaliprasad]

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

 

[Greg]

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 $$x>3$$.

 

[Prove It]

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...

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