# 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:

$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$

• I think this is the fastest. Oct 24, 2015 at 11:25
• Draw a graph with the lines $y = -4x -3$, $y = x+2$, $y = x-1$, and $y = x-3$. Then mark on the abscissa axis the intervals where $0,2$, or $4$ of them are negative.
– A.P.
Oct 24, 2015 at 11:33

## 5 Answers

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 $x \in \:\left(-2,-\frac{3}{4}\right)\cup \left(1,\:3\right)$

• My bad ,,, it should be factors. Oct 24, 2015 at 11:30
• The second factor is $-3/4$, not $3/4$. (Fortunately for the answer, correcting the sign doesn't affect the order of the factors.) Oct 24, 2015 at 11:47

\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}

$$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$$ $$\frac{(x+1)}{(x-1)(x+2)}-\frac{(x)}{(x-1)(x-3)}>0$$ $$\frac{1}{(x-1)}\left(\frac{(x+1)}{(x+2)}-\frac{(x)}{(x-3)}\right)>0$$ $$\frac{1}{(x-1)}\left(\frac{(x+1)(x-3)-x(x+2)}{(x+2)(x-3)}\right)>0$$ $$\frac{x^2-2x-3-x^2-2x}{(x-1)(x+2)(x-3)}>0$$ $$\frac{4x+3}{(x-1)(x+2)(x-3)}<0$$

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 $(-2,-\frac{3}{4})\cup(1,3)$

$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$

$\frac{(x+1)(x-3)}{(x-1)(x+2)(x-3)}-\frac{(x)(x+2)}{(x-1)(x+2)(x-3)}>0$

$\frac{(x+1)(x-3)-(x)(x+2)}{(x-1)(x+2)(x-3)}>0$

$\frac{4x+3}{(x-1)(x+2)(x-3)}<0$

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 $(-2,-\frac{3}{4})\cup(1,3)$

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: $$(-2,-\frac34)\cup(1,2).$$