# Question 4aac3

Feb 3, 2017

#### Explanation:

Make it an inequality with $0$ on one side, then do a sign analysis (sign chart/diagram/table -- whatever you're calling it).

The given inequality is equivalent to

(x − 1)/(x + 1) + (x)/(x − 2) - 1 - 10/((x − 2)(x + 1)) < 0#

Combine the left side into one quotient to get

$\frac{{x}^{2} - x - 6}{\left(x - 2\right) \left(x + 1\right)} < 0$.

This is equivalent to

$\frac{\left(x - 3\right) \left(x + 2\right)}{\left(x - 2\right) \left(x + 1\right)} < 0$

The partition numbers (or key numbers or whatever you're calling them -- some textbooks don't provide a name) are the zeros of the numerator and of the denominator. So they are

$- 2$, $- 1$, $2$, and $3$.

Analyze the sign of the quotient in the intervals determined by these numbers.

On $\left(- \infty , - 2\right)$, the quotient is positive

On $\left(- 2 , - 1\right)$, the quotient is negative

On $\left(- 1 , 2\right)$, the quotient is positive

On $\left(2 , 3\right)$, the quotient is negative

On $\left(3 , \infty\right)$, the quotient is positive.

We want the values of $x$ that make the quotient negative, so the solution set is

$\left(- 2 , - 1\right) \cup \left(2 , 3\right)$.