SOLUTION: 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

Algebra ->  Inequalities -> SOLUTION: 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      Log On


   



Question 895479: Problem regarding %282x%2B1%29%2F%283x-1%29+%3E+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+=+%282x%2B1%29%2F%283x-1%29 and highlighted the area. I got the answer 1/3 < x < 2
I solved it in this way too..
%282x%2B1%29%2F%283x-1%29+-+1+%3E+0 and got the same solution.
But my problem is... WHY CAN'T I CROSS MULTIPLY LIKE THIS???
2x%2B1+%3E+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.

Found 2 solutions by Fombitz, MathTherapy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You can solve it this way too.
%282x%2B1%29%2F%283x-1%29=1
2x%2B1=3x-1
-x=-2
x=-2
So this is one of the critical values.
The other is x=1%2F3 since the denominator goes to zero.
Now split the number line into three regions and test the inequality in those three regions.
.
.
.
A better way to say it is that by multiplying, you're losing information that's included when the value of the denominator crosses from one sign to the other and it's irretrievable because the assumption you make is that the denominator doesn't go to zero (but in fact it does).
If the denominator never goes to or through zero, you can solve it this way.
Additionally, allowing for division by zero leads to contradictions.
Hope that helps.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Problem regarding %282x%2B1%29%2F%283x-1%29+%3E+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+=+%282x%2B1%29%2F%283x-1%29 and highlighted the area. I got the answer 1/3 < x < 2
I solved it in this way too..
%282x%2B1%29%2F%283x-1%29+-+1+%3E+0 and got the same solution.
But my problem is... WHY CAN'T I CROSS MULTIPLY LIKE THIS???
2x+%2B+1+%3E+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.

%282x+%2B+1%29%2F%283x+-+1%29+%3E+1, with x+%3C%3E+1%2F3, since this value would render the inequality UNDEFINED
In this case, you certainly CAN cross-multiply %282x+%2B+1%29%2F%283x+-+1%29+%3E+1 to get: 2x+%2B+1+%3E+3x+-+1
2x+-+3x+%3E+-+1+-+1
-+x+%3E+-+2
x+%3C+%28-+2%29%2F-+1 ------ Note that the INEQUALITY sign changed, since DIVISION by a negative value (- 1) dictates such
Therefore, x+%3C+2
We now have 2 CRITICAL POINTS: 1%2F3, and 2
The intervals to test for values to make the inequality TRUE would then be:
1) x+%3C+1%2F3, and
2) 1%2F3+%3C+x+%3C+2
For x+%3C+1%2F3, we can use 0 for x
Does the INEQUALITY PROVE TRUE when this occurs? Let's see!!
With x = 0, %282x+%2B+1%29%2F%283x+-+1%29+%3E+1 becomes: %282%280%29+%2B+1%29%2F%283%280%29+-+1%29+%3E+1
%280+%2B+1%29%2F%280+-+1%29+%3E+1____1%2F-1+%3E+1____-+1+%3E+1 (FALSE)
For 1%2F3+%3C+x+%3C+2, we can use 1 for x
Does the INEQUALITY PROVE TRUE when this occurs? Let's see!!
With x = 1, %282x+%2B+1%29%2F%283x+-+1%29+%3E+1 becomes: %282%281%29+%2B+1%29%2F%283%281%29+-+1%29+%3E+1
%282+%2B+1%29%2F%283+-+1%29+%3E+1____3%2F2+%3E+1 (TRUE)
As seen from the 2 tested intervals, the only solution is: highlight_green%28highlight_green%281%2F3+%3C+x+%3C+2%29%29
***Another test interval is: x+%3E+2, but I knew that this interval would render the INEQUALITY, false.
So, your answer is CORRECT!! Good job!!
You can do the check!!
Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com.
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