Solving the Inequality $\frac{14x}{x+1}<\frac{9x-30}{x-4}$ The question says to find all the integral values of x for which the inequality holds.
the question is 
$$\frac{14x}{x+1}<\frac{9x-30}{x-4}$$
My Solution
\begin{align}
& \frac{14x}{x+1} < \frac{9x-30}{x-4} \\[6pt]
& \frac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}<0 \\[6pt]
& \frac{14x^2-64x-9x^2+21x+30}{(x+1)(x-4)} < 0 \\[6pt]
& \frac{5x^2-43x+30}{(x+1)(x-4)}<0
\end{align}
using quadratic formula, roots of $5x^2-43x+30$ comes $7.83$(approx.) and $0.763$(approx.)
so rewriting the equation as
$$\frac{(x-7.83)(x-0.763)}{(x+1)(x-4)}<0$$
I then did the plotting of zeroes and poles on number line for finding the values for $x$ but I donot get 2 integral values (which is the answer). Can anyone tell where I did wrong?
 A: Basic approach. Perhaps easier would be to rewrite the original inequality as
$$
14 - \frac{14}{x+1} < 9 + \frac{6}{x-4}
$$
This leads to
$$
\frac{14}{x+1} + \frac{6}{x-4} > 5
$$
which becomes
$$
\frac{4x-10}{(x+1)(x-4)} > 1
$$
You can now (a) consider the cases $x = 0, 1, 2, 3$ separately, and then otherwise, (b) for $x < -1$ or $x > 4$, we have
$$
(x+1)(x-4) < 4x-10
$$
which becomes
$$
x^2-7x+6 < 0
$$
which you should be able to handle.  Keep in mind that this inequality is only valid for the subcase (b) $x < -1$ or $x > 4$.
(There's probably a simpler way, incidentally.  See alans' comment, for instance.  This is just what I wrote up.)
A: Inequality is equivalent with $$\frac{(x-1)(x-6)}{(x+1)(x-4)}<0.$$ Function $f(x)=\frac{(x-1)(x-6)}{(x+1)(x-4)}$ has sign $+$ on intervals $(-\infty,-1)$, $(1,4)$, $(6,\infty)$ and sign $-$ on intervals $(-1,1)$ and $(4,6)$. 
Therefore, only integer solutions that satisfy inequality are $0$ and $5$.
A: One easy way to solve is to consider the corresponding equation.  A change of sings can only occur due to continuity at the equation's zeros or where the equation is undefined. 
A: Correcting your initial mistake will lead to the inequality
\begin{equation}
\dfrac{(x+1)(x-6)}{(x+1)(x-4)}<0
\end{equation}
which is positive on the interval $(6,\infty)$ since the product of $x$ coefficients is positive, and the expression changes sign at each zero and pole as you move left along the $x$-axis since each of the four linear factors is raised to an odd power. Thus it is negative only on the intervals $(-1,1)$ and $(4,6)$.
So the only integral solutions are $x=0$ and $x=5$.
