Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 !

share|improve this question
    
4  
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
1  
Have you tried drawing a graph of your function? –  Mark Bennet Oct 5 '11 at 14:16
1  
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

1 Answer 1

up vote 2 down vote accepted

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. $-5 \lt x \le 4$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.