# Math Help - Interval Form?

1. ## Interval Form?

how do i write this in interval form??

(x+2)/(x-5)≤0

2. Originally Posted by VNVeteran
how do i write this in interval form??

(x+2)/(x-5)≤0
You should first note that $x \neq 5$.

$\frac{x + 2}{x - 5} \leq 0$

$1 + \frac{7}{x - 5} \leq 0$

$\frac{7}{x - 5} \leq 1$

Now we will have two alternatives, depending on whether the denominator is positive or negative...

Case 1: $x - 5$ is positive, which means $x > 5$.

$7 \leq x - 5$

$14 \leq x$

$x \geq 14$.

Case 2: $x - 5$ is negative, which means $x < 5$.

$7 \geq x - 5$

$14 \geq x$

$x \leq 14$.

It's pretty obvious that the ONLY $x$ which can not satisfy this equation are $x = 5$.

So the domain is $x < 5 \cup x > 5$.

Alternatively, it can be written $x \neq 5$ or $(-\infty , 5) \cup (5, \infty)$.

3. Originally Posted by VNVeteran
how do i write this in interval form??

(x+2)/(x-5)≤0
please be specific. are we talking about writing the domain in interval notation or the solution to the inequality in interval notation?

I suspect the latter. In which case set the numerator and denominator to zero. You get x = -2 or x = 5 (now note that x cannot be 5, because the denominator cannot in fact be zero, we just do this to figure out the signs). Now you can draw a number line and plot x = -2 and x = 5 on it. Then plug in numbers in each of the three intervals that you have. you will find that the middle interval is the only one that works. the resulting solution is $[-2, 5)$

4. Originally Posted by VNVeteran
how do i write this in interval form??

(x+2)/(x-5)≤0
A fraction is negative if and only if numerator and denominator are of different signs. That is:

1) $x+2\ge 0$ and x- 5< 0
which lead to $x\ge -2$ and x< 5: [-2, 5)
or

2) $x+2\le 0$ and x- 5> 0
which lead to $x\le -2$ and x> 5 which are impossible together.

5. Originally Posted by Prove It
You should first note that $x \neq 5$.

$\frac{x + 2}{x - 5} \leq 0$

$1 + \frac{7}{x - 5} \leq 0$

$\frac{7}{x - 5} \leq 1$

Now we will have two alternatives, depending on whether the denominator is positive or negative...

Case 1: $x - 5$ is positive, which means $x > 5$.

$7 \leq x - 5$

$14 \leq x$

$x \geq 14$.

Case 2: $x - 5$ is negative, which means $x < 5$.

$7 \geq x - 5$

$14 \geq x$

$x \leq 14$.

It's pretty obvious that the ONLY $x$ which can not satisfy this equation are $x = 5$.

So the domain is $x < 5 \cup x > 5$.

Alternatively, it can be written $x \neq 5$ or $(-\infty , 5) \cup (5, \infty)$.
ok do i put any brackets in the interval form????? thats mainly what im confused about. when do i use brackets and when do i not?

6. Originally Posted by HallsofIvy
A fraction is negative if and only if numerator and denominator are of different signs. That is:

1) $x+2\ge 0$ and x- 5< 0
which lead to $x\ge -2$ and x< 5: [-2, 5)
or

2) $x+2\le 0$ and x- 5> 0
which lead to $x\le -2$ and x> 5 which are impossible together.
no disrespect but i think your wrong. my book says the answer is (-infin, 5] U [5, infin)

7. no, he's correct, and i do think you should reconsider twice on determining when an answer is wrong or not, see the procedure first.

i wouldn't be surprised of another typo on a book, so in this case, book's answer is wrong and HallsofIvy is correct.

8. HallsofIvy's answer is indeed correct. To demonstrate why $(-\infty,5)\cup(5,\infty)$ is wrong, try substituting $x=6$ into the inequality and see what you get. Actually, any $x>5$ gives a positive result, in contradiction to this answer.

9. hello
if we take the first case of what "Hallsofivy" posted.
we'll have $-2\leq x< 5$ as a solution.

10. Originally Posted by Raoh
hello
if we take the first case of what "Hallsofivy" posted.
we'll have $-2\leq x< 5$ as a solution.
However, IF the original question was "what is the domain of the function $\frac{x+2}{x- 5}$, THEN the answer is $(-\infty, 5)\cup(5, \infty)$. Prove It said that in his first response!

However, I don't see what the "<=" has to do with it then. You find domains of functions, not inequalities.