Question 895479
Problem regarding {{{(2x+1)/(3x-1) > 1}}}

I actually know how to solve it. I solved in many ways because I wanted to be an expert of it.

I plotted a graph taking {{{y = (2x+1)/(3x-1)}}} and highlighted the area. I got the answer 1/3 < x < 2

I solved it in this way too..

{{{(2x+1)/(3x-1) - 1 > 0}}} and got the same solution.

But my problem is... WHY CAN'T I CROSS MULTIPLY LIKE THIS???

{{{2x + 1 > 3x - 1}}} 

Please don't tell "Because we don't know whether 'x' is plus or minus which changes > to <   I want a better explanation. Thank you so much.
<pre>
{{{(2x + 1)/(3x - 1) > 1}}}, with {{{x <> 1/3}}}, since this value would render the inequality UNDEFINED
In this case, you certainly CAN cross-multiply {{{(2x + 1)/(3x - 1) > 1}}} to get: {{{2x + 1 > 3x - 1}}} 
{{{2x - 3x > - 1 - 1}}}
{{{- x > - 2}}}
{{{x < (- 2)/- 1}}} ------ Note that the INEQUALITY sign changed, since DIVISION by a negative value (- 1) dictates such
Therefore, {{{x < 2}}}
We now have 2 CRITICAL POINTS: {{{1/3}}}, and {{{2}}}
The intervals to test for values to make the inequality TRUE would then be:
1) {{{x < 1/3}}}, and
2) {{{1/3 < x < 2}}}

For {{{x < 1/3}}}, we can use 0 for x
Does the INEQUALITY PROVE TRUE when this occurs? Let's see!!
With x = 0, {{{(2x + 1)/(3x - 1) > 1}}} becomes: {{{(2(0) + 1)/(3(0) - 1) > 1}}}
{{{(0 + 1)/(0 - 1) > 1}}}____{{{1/-1 > 1}}}____{{{- 1 > 1}}} (FALSE)

For {{{1/3 < x < 2}}}, we can use 1 for x
Does the INEQUALITY PROVE TRUE when this occurs? Let's see!!
With x = 1, {{{(2x + 1)/(3x - 1) > 1}}} becomes: {{{(2(1) + 1)/(3(1) - 1) > 1}}}
{{{(2 + 1)/(3 - 1) > 1}}}____{{{3/2 > 1}}} (TRUE)
As seen from the 2 tested intervals, the only solution is: {{{highlight_green(highlight_green(1/3 < x < 2))}}}
***Another test interval is: {{{x > 2}}}, but I knew that this interval would render the INEQUALITY, false.
So, your answer is CORRECT!! Good job!!
You can do the check!! 

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