How do you solve #(x+2)/(x+1)>0#?

1 Answer
Nov 17, 2016

Answer:

The "critical" values are #x=-2# and #x=-1#. When #x<-2#, we get #-/-##=+#. When #-2< x<-1#, we get #+/-##=-#. When #x> -1#, we get #+/+##=+#. Thus, our solution is #x in (-oo,-2)uu(-1,oo).#

Explanation:

In order for the fraction as a whole to be greater than 0, both the numerator and the denominator will need to be greater than 0 (or less than 0), so that the division will yield a positive value.

We can think of the LHS as the product of two factors: #x+2# and #1/(x+1)#. That is,

#(x+2)(1/(x+1))>0#

The first factor #x+2# will be greater than 0 when x is greater than -2:

#x+2>0#
#<=>x> -2#

The second factor #1/(x+1)# is a little more tricky, but it's easy when you realize that a fraction like this will be positive only when its denominator is the same sign as the numerator (in this case, positive). The numerator is a constant; it cannot affect the fraction's sign. In other words:

#1/(x+1)>0#
#<=>x+1>0#
#<=>x> -1#

Next, we create a sign table for the factors and fill it in with #+# or#-# signs depending on where each factor is positive or negative. Our "points of interest" are -1 and -2.

Factor - - - - - - - - - - - Sign - - - - - - -
. . . . . . . . .------- -2 ---------- -1 -----------
#x+2# . . . #--0+++++#
#x+1# . . . #-----O/++#
. . . . . . . . .-------------------------------------
#(x+2)/(x+1)# . . #++0---O/++#

To fill in the bottom row, simply multiply the signs of all the factor rows together.

Notice the #O/# for the factor #x+1#. That's there because #x+1# is a denominator factor, and any #x#-value that makes such a factor equal to 0 will make the whole expression undefined.

Finally, we pluck out the interval(s) over which the whole expression is greater than 0. After multiplying the signs in the sign table, we see that we get:

#(x+2)/(x+1)>0# when #x<-2# and #x> -1#.

Our solution is: #{x|x<-2uux> -1}#, or
#x in (-oo,-2)uu(-1,oo)#.