Question

Question #4aac3

Answer 1

Please see below.

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

`(x^2-x-6)/((x-2)(x+1)) < 0`.

This is equivalent to

`((x-3)(x+2))/((x-2)(x+1)) < 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 `(-oo, -2)`, the quotient is positive

On `(-2, -1)`, the quotient is negative

On `(-1, 2)`, the quotient is positive

On `(2, 3)`, the quotient is negative

On `(3, oo)`, the quotient is positive.

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

`(-2,-1) uu (2,3)`.