Question #4aac3

1 Answer
Jim H
Feb 3, 2017

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)#.

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