# Find the domain and the solution

I am trying to understand the solution of this problem.

$$\dfrac{(2x-8)}{(x+5)} \leq0$$

The domain is $\mathbb{R}\ - \{-5\}$ and the solution is $-5 \lt x \leq 4$ However, I can't figure why $-5$ is not included in the solution. I did a table of sign :

x | -10 | -5 | 0 | 4 | 10
|   + |  0 | - | E | +


It is "$\leq 0$" so I select everything that is smaller or equals to 0...

$x=-5, 2*-5-8/-5+5 = 0 .. 0$ is equal to $0$, why is it excluded in the solution ?

Thanks !

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–  Chris Godsil Oct 5 '11 at 13:12
You cannot divide by zero. –  lhf Oct 5 '11 at 13:13
What is the E in your sign table? Shouldn't it be 0? –  Ross Millikan Oct 5 '11 at 13:51
Have you tried drawing a graph of your function? –  Mark Bennet Oct 5 '11 at 14:16
In the table, the entry for $x=-5$ should not say $0$, it should say something ND (not defined), or E, which is what my calculator says. And the entry for $x=4$ should be $0$. Maybe you just transposed? –  André Nicolas Oct 5 '11 at 16:10

In the equation $$\dfrac{(2x-8)}{(x+5)} \leq0$$ if you put $x = -5$ it reduces to $$\frac{-18}{0} \le 0$$ as the denominator is zero the left side of this inequality is known as division by zero.From the same Wikipedia url :

Whether this expression can be assigned a well-defined value depends upon the mathematical setting. In ordinary (real number) arithmetic, the expression has no meaning, as there is no number which, multiplied by 0, gives a (a≠0).

So as we are dealing with real number arithmetic here we are excluding $x=-5$ from the domain.

Coming to the solution, $$\dfrac{(2x-8)}{(x+5)} \leq0 \Rightarrow (2x-8) \leq0\Rightarrow x \leq 4$$ only if we assume $(x+5) \neq 0$,but then either $(x+5) \gt 0$ or $(x+5) \lt 0$.

But when $(x+5) \lt 0$,multiplying will reverse the inequality giving $(2x-8) \geq0$ hence,$x \lt -5$ and $x \ge 4$ which makes no sense,hence discarded.

So the only possible solution is when $x \gt -5$ and $x \lt 4$ i,e. $-5 \lt x \le 4$

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