# Solve the inequity

• Sep 21st 2009, 04:20 PM
B-lap
Solve the inequity
Solve the following inequity. Write the anwser in interval notation.

1 + 2/x+1 greater or less than or equal to 2/x

i got the intervals (-infinity, -2) U [-2, 1] U [1, infinity)

Are these the correct intervals, and which ones are true?
• Sep 21st 2009, 04:23 PM
skeeter
Quote:

Originally Posted by B-lap
Solve the following inequity. Write the anwser in interval notation.

1 + 2/x+1 greater or less than or equal to 2/x

i got the intervals (-infinity, -2) U [-2, 1] U [1, infinity)

Are these the correct intervals, and which ones are true?

the bold statement makes no sense ... fix it.
• Sep 21st 2009, 04:23 PM
B-lap
less than or equal to
• Sep 21st 2009, 04:46 PM
skeeter
Quote:

Originally Posted by B-lap
Solve the following inequity. Write the anwser in interval notation.

1 + 2/x+1 less than or equal to 2/x

i got the intervals (-infinity, -2) U [-2, 1] U [1, infinity)

Are these the correct intervals ... no

$\displaystyle 1 + \frac{2}{x+1} - \frac{2}{x} \le 0$

$\displaystyle \frac{x(x+1)}{x(x+1)} + \frac{2x}{x(x+1)} - \frac{2(x+1)}{x(x+1)} \le 0$

$\displaystyle \frac{x^2 + x + 2x - 2x - 2}{x(x+1)} \le 0$

$\displaystyle \frac{x^2 + x - 2}{x(x+1)} \le 0$

$\displaystyle \frac{(x+2)(x-1)}{x(x+1)} \le 0$

critical values are $\displaystyle x = -2$, $\displaystyle x = -1$, $\displaystyle x = 0$ , and $\displaystyle x = 1$

now check the inequality using x-values within each interval separated by the critical values of x.