Question 300795
{{{(x)/(x+20)>(2)/(x+8)}}}
Look at the inequality and determine when the left hand side equals the right hand side. 
{{{ (x)/(x+20)=(2)/(x+8)}}}
{{{x(x+8)=2(x+20)}}}
{{{x^2+8x=2x+40}}}
{{{x^2+6x-40=0}}}
You can factor this quadratic,
{{{(x+10)(x-4)=0}}}
Two solutions: {{{x=-10}}} and {{{x=4}}}
In addition, because of the x terms in the denominator, add the additional points when the denominator goes to zero. 
{{{x=-20}}} and {{{x=-8}}}
Set up a number line and break it up using these two points.
Region 1: ({{{-infinity}}},{{{ -20}}})
 Region 2: ({{{-20}}},{{{ -10}}})
Region 3: ({{{-10}}},{{{ -8}}})
Region 4: ({{{-8}}},{{{4}}})
Region 5: ({{{4}}},{{{ infinity}}})
Choose points in each region and check the inequality in that region to see if it's valid or not. 

Region 1: Pick {{{x=-30}}}.
{{{(x)/(x+20)>(2)/(x+8)}}}
{{{(-30)/(-30+20)>(2)/(-30+8)}}}
{{{-30/-10>2/-22}}}
{{{3>-1/11}}}
True, Region 1 is a valid region.
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Region 2: Pick {{{x=-15}}}.
{{{(-15)/(-15+20)>(2)/(-15+8)}}}
{{{(-15)/(5)>(2)/(-7)}}}
{{{-3>-(2/7)}}}
False, Region 2 is not a valid region.
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Region 3: Pick x=-9.
{{{(x)/(x+20)>(2)/(x+8)}}}
{{{(-9)/(-9+20)>(2)/(-9+8)}}}
{{{9/11>-2}}}
True, Region 3 is  a valid region.
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Region 4: Pick x=0.
{{{(x)/(x+20)>(2)/(x+8)}}}
{{{0>(2)/(0+8)}}}
{{{0>1/4}}}
False, Region 4 is not a valid region.
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Region 5: Pick x=10.
{{{(x)/(x+20)>(2)/(x+8)}}}
{{{10/30>2/18}}}
{{{1/3>1/9}}}
True, Region 5 is a valid region.
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({{{-infinity}}},{{{ -20}}})U({{{-10}}},{{{ -8}}})U({{{4}}},{{{ infinity}}})
E is the correct answer.