What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
 A: Clearly
$\frac{x+1}{\left(x-1\right)\left(x+2\right)}-\frac{x}{\left(x-1\right)\left(x-3\right)}>0$
$\Rightarrow \frac{-4x-3}{\left(x+2\right)\left(x-1\right)\left(x-3\right)}>0$
Then draw a straight line and mark $x$ values that make the fractors zero. That is $-2 ,-\frac{3}{4} ,1,3$.
Then check the validity for some $x$ value less than $-2$. 
Say $x=-3$.
For $x=-3$ we have that inequality does not hold.
So LHS is negative for $x=-3$
Then clearly $x \in \:\left(-2,-\frac{3}{4}\right)\cup \left(1,\:3\right)$
A: $$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$$
$$\frac{(x+1)}{(x-1)(x+2)}-\frac{(x)}{(x-1)(x-3)}>0$$
$$\frac{1}{(x-1)}\left(\frac{(x+1)}{(x+2)}-\frac{(x)}{(x-3)}\right)>0$$
$$\frac{1}{(x-1)}\left(\frac{(x+1)(x-3)-x(x+2)}{(x+2)(x-3)}\right)>0$$
$$\frac{x^2-2x-3-x^2-2x}{(x-1)(x+2)(x-3)}>0$$
$$\frac{4x+3}{(x-1)(x+2)(x-3)}<0$$
The above inequality implies the following:
$$x\not = 1,x\not = -2,x\not = 3,x\not = -\frac{3}{4}$$
Now check the values of $x$ in the following intervals: $(-\infty,-2),(-2,-\frac{3}{4}),(-\frac{3}{4},1),(1,3),(3,+\infty)$ 
You will find that the inequality holds in the interval $(-2,-\frac{3}{4})\cup(1,3)$
A: $$\small\begin{align}&\frac{x+1}{(x-1)(x+2)}\gt\frac{x}{(x-1)(x-3)}\\\\&\iff \frac{x+1}{(x-1)(x+2)}\cdot (x-1)^2(x+2)^2(x-3)^2\gt\frac{x}{(x-1)(x-3)}\cdot (x-1)^2(x+2)^2(x-3)^2\\\\&\iff (x+1)(x-1)(x+2)(x-3)^2\gt x(x-1)(x+2)^2(x-3)\\\\&\iff (x-1)(x+2)(x-3)((x+1)(x-3)-x(x+2))\gt 0\\\\&\iff (x-1)(x+2)(x-3)(-4x-3)\gt 0\\\\&\iff (x-1)(x+2)(x-3)(4x+3)\lt 0\\\\&\iff -2\lt x\lt -\frac 34\quad\text{or}\quad 1\lt x\lt 3\end{align}$$
A: $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
$\frac{(x+1)(x-3)}{(x-1)(x+2)(x-3)}-\frac{(x)(x+2)}{(x-1)(x+2)(x-3)}>0$
$\frac{(x+1)(x-3)-(x)(x+2)}{(x-1)(x+2)(x-3)}>0$
$\frac{4x+3}{(x-1)(x+2)(x-3)}<0$
the inequations are:


*

*$x>-2$

*$4x+3>0$ or $x>-\frac{4}{3}$

*$x>1$

*$x>3$


We need to have one or three inequations true to have the big inequality hold.
You will find that the answer is  $(-2,-\frac{3}{4})\cup(1,3)$
A: Reducing to the same denominator, we get
$$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}\iff -\frac{4x+3}{(x+2)(x-1)(x-3)}>0 $$
Now the sign of $-\dfrac{4x+3}{(x+2)(x-1)(x-3)}$, when defined, is the sign of the product $p(x)=-(4x+3)(x+2)(x-1)(x-3)$, which by Bolzano's theorem can change sign only at $-2, -\dfrac34,1$ and $3$. As $p(0)=-\frac12<0$, we only have to alternate signs to obtain the following table:
$$\begin{array}{*{9}{c}}
&-2&&-\frac34&&1&&2& \\
\hline
-&\Vert&  + &0&\color{red}{\mathbf-}&\Vert&+&\Vert&-\end{array}, $$
whence the solutions:
$$(-2,-\frac34)\cup(1,2).$$
