Question 124605
<pre><font face = "courier new" size = 4><b>
Can someone please help me a.s.a.p?
The solutions to {{{1/(x+2)-2/(x-1)>0}}}
My answer options are:
a.(-infinity,5)U(1,infinity)
b.(-infinity,-5)U(-2,1)
c.(-5,-2)U(1,5)
d.(-2,1)U(5,infinity)
I thought the answer was b., what do you think? Thanks


{{{1/(x+2)-2/(x-1)>0}}}

Get LCD on the left:

{{{((x-1)-2(x+2))/((x+2)(x-1))>0}}}

{{{(x-1-2x-4)/((x+2)(x-1))>0}}}

{{{(-x-5)/((x+2)(x-1))>0}}}

To find the critical values, set each factor
of the numerator and the denominator = 0, and
solve for x:

{{{-x-5=0}}} yields critical value x = -5

{{{x+2=0}}} yields critical value x = -2

{{{x-1=0}}} yields critical value x = 1

Mark the critical values on a number line:

-------o--------o--------o------
-7 -6 -5 -4 -3 -2 -1  0  1  2  3

Pick any value left of -5, say -6.
Substitute it into:

{{{(-x-5)/((x+2)(x-1))>0}}}
{{{(-(-6)-5)/((-6+2)(-6-1))>0}}}
{{{(6-5)/((-4)(-7))>0}}}
{{{1/28>0}}}

That is true, so shade the part of the number
line to the left of -5.

<======o--------o--------o------
-7 -6 -5 -4 -3 -2 -1  0  1  2  3

Next, pick any value between of -5 and -2, say -3.
Substitute it into:

{{{(-x-5)/((x+2)(x-1))>0}}}
{{{(-(-3)-5)/((-3+2)(-3-1))>0}}}
{{{(3-5)/((-1)(-4))>0}}}
{{{(-2)/4>0}}}
{{{-1/2>0}}}

That is false, so do not shade the part of the number
line between -5 and -3, so we still have:

<======o--------o--------o------
-7 -6 -5 -4 -3 -2 -1  0  1  2  3  

Pick any value between -2 and 1, say 0.
Substitute it into:

{{{(-x-5)/((x+2)(x-1))>0}}}
{{{(-(0)-5)/((0+2)(0-1))>0}}}
{{{(0-5)/((2)(-1))>0}}}
{{{(-5)/(-2)>0}}}
{{{5/2>0}}}

That is true, so shade the part of the number
line between -2 and 1.

<======o--------o========o------
-7 -6 -5 -4 -3 -2 -1  0  1  2  3

Pick any value right of 1, say 2.
Substitute it into:

{{{(-x-5)/((x+2)(x-1))>0}}}
{{{(-(2)-5)/((2+2)(2-1))>0}}}
{{{(-2-5)/((4)(1))>0}}}
{{{-7/4>0}}}

That is false, so do not shade the part of the number
line to the right of 1.

So far the number line so far gives: 

<======o--------o========o------
-7 -6 -5 -4 -3 -2 -1  0  1  2  3

We are technically supposed to test the critical values
themselves to see if they are solutions.  But none do
since substituting -5 or -1 for x gives 0 > 0, and
substituting 1 for x causes the left side to be undefined.

So that number line is represented by this interval notation: 

(-oo,-5) U (-2,1)

So you are correct since that's choice b.

Edwin</pre>