Question

PLEASE HELP ILL GIVE MEDALS AND MARK BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!! NEEDS TO BE ALEGABRA 2 MEATHOD!!!!!!!!
Solve x/x+1 < x/x-1

Answer 1

Hi there,
This is the original inequality equation:
$\frac{x}{x+1} \ \textless \ \frac{x}{x-1}$
So, we first need to find the critical points of equality, and we can do that by switching the less than sign to an equal sign.
$\frac{x}{x+1} = \frac{x}{x-1}$
Now, we multiply both sides by x + 1:
$x= \frac{x^{2} +x}{x-1}$
Then, we multiply both sides by x - 1:
$x^{2} -x= x^{2} +x$
Next, we subtract x² from both sides:
$-x=x$
After that, we solve for x. We do this by adding -x to both sides and dividing by 2. Doing so gives us x = 0, which is our first critical point. We need to find a few more critical points by testing x = -1 and x = 1. Here is how we do that:
x = −1 (Makes left denominator equal to 0)x = 1   (Makes right denominator equal to 0)Check intervals in between critical points. (Test values in the intervals to see if they work.)x <−1    (Doesn't work in original inequality)−1 < x <0  (Works in original inequality)0 < x < 1 (Doesn't work in original inequality)x > 1  (Works in original inequality)
Therefore, the answer to your query is -1 < x < 0 or x > 1. Hope this helps and have a phenomenal day!