Question 988807

Solve for x in the inequality:
-1 is less than (x+2)/(x-3)


Any tips, tricks, advice for other inequality problems is greatly appreciated as well as help with this problem! Thank you in advance!
<pre>{{{- 1 < (x + 2)/(x - 3)}}}, with {{{x <> 3}}} as this would make the denominator 0, and therefore, an UNDEFINED inequality 
{{{- 1(x - 3) < x + 2}}} ----- Multiplying both sides by LCD, x - 3
{{{- x + 3 < x + 2}}}
{{{- x < x + 2 - 3}}} -------- Subtracting 3 from both sides
{{{- x < x - 1}}}
{{{- x - x < - 1}}} --------- Subtracting x from both sides
{{{- 2x < - 1}}}
{{{(- 2x)/(- 2) > (- 1)/(- 2)}}} -------- Inequality sign changes when dividing by a negative value
{{{x > 1/2}}}
Now, we have the CRITICAL VALUES: {{{system((1/2)_and,3)}}} 
This means that we will have THREE (3) TEST-INTERVALS to determine whether or not values
in the intervals SATISFY the ORIGINAL inequality: {{{- 1 < (x + 2)/(x - 3)}}}. These TEST INTERVALS are:
1)   {{{x < 1/2}}}
2)   {{{1/2 < x < 3}}}, and
3)   {{{x > 3}}}

Testing {{{x < 1/2}}} with test value: {{{x = 0}}}, {{{- 1 < (x + 2)/(x - 3)}}} becomes: {{{- 1 < (0 + 2)/(0 - 3)}}}
{{{- 1 < 2/(- 3)}}}
{{{- 1 < - 2/3}}}
As the above is TRUE, the test-interval: {{{highlight_green(x < 1/2)}}} IS a solution

Testing {{{1/2 < x < 3}}} with test value: {{{x = 1}}}, {{{- 1 < (x + 2)/(x - 3)}}} becomes: {{{- 1 < (1 + 2)/(1 - 3)}}}
{{{- 1 < 3/(- 2)}}}
{{{- 1 < - 1&1/2}}}
As the above is FALSE, the test-interval: {{{1/2 < x < 3}}} IS NOT a solution

Testing {{{x > 3}}} with test value: {{{x = 4}}}, {{{- 1 < (x + 2)/(x - 3)}}} becomes: {{{- 1 < (4 + 2)/(4 - 3)}}}
{{{- 1 < 6/1}}}
{{{- 1 < 6}}}
As the above is TRUE, the test-interval: {{{highlight_green(x > 3)}}} IS a solution
In interval notation, this is: <font face = "Tohoma" size = 4 color = "indigo"><b>(- oo, ½) &#5196 (3, oo) </font face = "Tohoma" size = 4 color = "indigo"></b>, or ({{{- infinity}}}{{{","}}}{{{1/2}}}) &#5196 ({{{3}}}{{{","}}}{{{infinity}}})