Question 1199000
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As the tutor @greenestamps has shown
{{{(x+9)/(x+1) -2 > 0}}}
simplifies to
{{{(-x+7)/(x+1) > 0}}}


If x = 7, then the numerator is 0.
If x = -1, then the denominator is 0, so we must avoid this x value.


Draw a number line with -1 and 7 marked on it.
Label the following regions or intervals
P: -infinity < x < -1
Q: -1 < x < 7
R: 7 < x < infinity
{{{drawing(400,200,-4,9,-2,2,

line(-4-2,0,9+2,0),
line(-4,0.1,-4,-0.1),
line(-3,0.1,-3,-0.1),
line(-2,0.1,-2,-0.1),
line(-1,0.1,-1,-0.1),
line(0,0.1,0,-0.1),
line(1,0.1,1,-0.1),
line(2,0.1,2,-0.1),
line(3,0.1,3,-0.1),
line(4,0.1,4,-0.1),
line(5,0.1,5,-0.1),
line(6,0.1,6,-0.1),
line(7,0.1,7,-0.1),
line(8,0.1,8,-0.1),
line(9,0.1,9,-0.1),
locate(-4-0.2,-0.1,"-4"),
locate(-3-0.2,-0.1,"-3"),
locate(-2-0.2,-0.1,"-2"),
locate(-1-0.2,-0.1,"-1"),
locate(0-0.2,-0.1,"0"),
locate(1-0.2,-0.1,"1"),
locate(2-0.2,-0.1,"2"),
locate(3-0.2,-0.1,"3"),
locate(4-0.2,-0.1,"4"),
locate(5-0.2,-0.1,"5"),
locate(6-0.2,-0.1,"6"),
locate(7-0.2,-0.1,"7"),
locate(8-0.2,-0.1,"8"),
locate(9-0.2,-0.1,"9"),

circle(-1,0,0.25),
circle(-1,0,0.27),

circle(7,0,0.25),
circle(7,0,0.27),

red(line(-1-0.27,0.02,-4,0.02)),
red(line(-1-0.27,0,-4,0)),
red(line(-1-0.27,-0.02,-4,-0.02)),
red(locate(-2.5,0.5,"P")),

blue(line(-1+0.27,0.02,7-0.27,0.02)),
blue(line(-1+0.27,0,7-0.27,0)),
blue(line(-1+0.27,-0.02,7-0.27,-0.02)),
blue(locate(3,0.5,"Q")),

green(line(7+0.27,0.02,10,0.02)),
green(line(7+0.27,0,10,0)),
green(line(7+0.27,-0.02,10,-0.02)),
green(locate(8.2,0.5,"R"))

)}}}
Note the open holes at -1 and 7.


Let's select an x value from interval P (shown in red above)
I'll pick x = -2
{{{(-x+7)/(x+1) > 0}}}


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


{{{(9)/(-1) > 0}}}


{{{-9 > 0}}}
which is false, so x = -2 is not a solution to the inequality.
By extension any item in interval P will also be non-solutions.
This rules out interval P from the solution set.


Now pick something from interval Q. 
I'll go for x = 0
{{{(-x+7)/(x+1) > 0}}}


{{{(-0+7)/(0+1) > 0}}}


{{{(7)/(1) > 0}}}


{{{7 > 0}}}
which is true. 
You should find similar true results for any other value in interval Q.
So interval Q is part of the answer.


Lastly, we need to check interval R.
Let's say we select x = 8
{{{(-x+7)/(x+1) > 0}}}


{{{(-8+7)/(8+1) > 0}}}


{{{(-1)/(9) > 0}}}


{{{-0.111 > 0}}} approximately
Like with interval P, we get a false statement
So anything in interval R leads to a false statement.



Therefore, the only valid solution set is -1 < x < 7
Which in interval notation would be (-1,7)
Be sure not to mix this up with ordered pair notation.


Confirmation can be done using a graph.
Desmos or GeoGebra are good free graphing tools.
{{{drawing(400,400,-10,10,-15,15,
graph(400,400,-10,10,-15,15,-100,(x+9)/(x+1) -2)
)}}}
The green curve represents y = (x+9)/(x+1) -2 aka y = (-x+7)/(x+1)
This green curve is above the x axis on the interval -1 < x < 7
It is below the x axis when either x < -1 or when x > 7.
The x intercept is 7, and there's a vertical asymptote at x = -1.


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Answer: <font color=red>Choice A) -1 < x < 7</font>
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