Find the set of the real numbers $x$ satisfying the given inequality $\frac{1}{x-4}<\frac{5}{x+1}$

Question

Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:

$\frac{1}{x-4}<\frac{5}{x+1}$

We know that $\frac{1}{x-4}<\frac{5}{x+1} \iff \frac{1}{x-4}-\frac{5}{x+1}<0$

$\iff \frac{x+1-5x+20}{(x-4)(x+1)}<0$

$\iff \frac{21-4x}{(x-4)(x+1)}<0$

$\frac{21-4x}{(x-4)(x+1)}$ will be negative, if

\begin{description} $21-4x<0$ and $(x-4)(x+1)>0$.

or,

$21-4x>0$ and $(x-4)(x+1)<0$

The first case,

$21-4x<0 \iff 4x>21 \iff x>\frac{21}{4}$

$(x-4)(x+1)>0$, if $(x-4)$ and $(x+1)$ have the same sign, this means that:

$x-4<0$ and $x+1<0$, then $x<4$ and $x<-1$

Thus, $\forall x \in (-\infty,4)\cap (-\infty,-1)=(-\infty,-1)$, $(x-4)(x+1)>0$

$x-4>0$ and $x+1>0$, then $x>4$ and $x>-1$.

Thus, $\forall x \in (4,\infty)\cap (-1,\infty)=(4,\infty)$, $(x-4)(x+1)>0$

SO, $\forall x \in (-\infty,-1) \cup (4,\infty)$, $(x-4)(x+1)>0$

Therefore, $\forall x \in (\frac{21}{4},\infty) \cap [(-\infty,-1) \cup (4,\infty)]=(\frac{21}{4},\infty)$, $\frac{1}{x-4}<\frac{5}{x+1}$.

The second case,

$21-4x>0 \iff 4x<21 \iff x<\frac{21}{4}$

$(x-4)(x+1)<0$, if $(x-4)$ and $(x+1)$ have a different signs, this means that:

$x-4<0$ and $x+1>0$, then $x<4$ and $x>-1$

Thus, $\forall x \in (-1,4)$, $(x-4)(x+1)<0$.

$x-4>0$ and $x+1<0$, then $x>4$ and $x<-1$, which is impossible.

Therefore, $\forall x \in (-\infty,-\frac{21}{4})\cup (-1,4)=(-1,4)$, $\frac{1}{x-4}<\frac{5}{x+1}$.

As a result, the solution set is $(\frac{21}{4},\infty) \cup (-1,4)$.

Is that true, please?

Answer 1

Yes, you are right, but after writing $$\frac{4x-21}{(x-4)(x+1)}>0$$ we get the answer immediately by the intervals method: $$(-1,4)\cup\left(\frac{21}{4},+\infty\right).$$

The intervals method here it's the following.

We need to draw a $x$ axis and to put points $-1$, $4$ and $\frac{21}{4}$.

Now, easy easy to see that the sing of $\frac{4x-21}{(x-4)(x+1)}$ for $x>\frac{21}{4}$ is $+$

and since degree of our points are odd (they are equal to $1$,

we see the the sing of the expression is changed.

Id est, we got the following sings on segments $(-\infty,-1),$ $(-1,4),$ $\left(4,\frac{21}{4}\right)$ and $\left(\frac{21}{4},+\infty\right)$. $$-,+,-,+$$ and we can write the answer.

Answer 2

Here's a shorter way, with less computations:

• If $x<-1$, both sides of the inequality are negative, so that $$\frac{1}{x-4}<\frac{5}{x+1}\iff x-4>\frac{x+1}5\iff5x-20>x+1\iff x>\frac{21}5,$$ which is impossible since $x<-4$.
• If $-1<x<4$, the l.h.s. $<0$ whereas the r.h.s. is $>0$, so the inequality is satisfied for all $x\in (-1,4)$.
• If $x>0$, both sides are positive, so, as in the first case, we obtain $x>\smash{\dfrac{21}5}$, condition which is not incompatible with this case.

To sum it up the solutions are $$(-1,4)\cup\Bigl(\frac{21}5,+\infty\Bigr).$$