1. Rational, absolute value equality

Hello there!
I have been struggling with the next equality for some time now. The problem is, that I can get to the correct answers, but they are in the wrong intervals, and I can not figure out, what is it, that I am missing.

So:
| (3x-2)/(x-1) | = 2

I understand, that the root is 2/3, and that if x>= 2/3, the expression inside the absolute value sign is positive, and negative otherwise. After taking the different cases into consideration, I get that if x is the element of (-inf,2/3), the solution is 4/5, and if x is the element of [2/3,inf), the solution is 0. I simply don't know, what am I doing wrong, I have had no problems, solving absolute value equations before, but this rational one confuses me a bit.

2. This is not an inequality...

3. Yes, you are right, I have mistyped it. Any suggestions concerning my problem?

4. Originally Posted by szucslaszlo
...
So:
$\begin{vmatrix}
\frac{3x-2}{x-1}
\end{vmatrix} = 2$

I understand, that the root is 2/3, and that if x>= 2/3, the expression inside the absolute value sign is positive, and negative otherwise. ...
First, the sign of the "inside" of the absolute value depends on the

3x - 2 (making x = 2/3 an "interesting point") AND on
x - 1 (making x = 1 an "interesting point").

You might want to make a number line, and mark the points x = 2/3 and x = 1. ...

5. Ah, yes, I understand. I forgot to inspect the two polinomials seperatly. I'll check back if it worked out his way. Thanks.

6. Originally Posted by szucslaszlo
Hello there!
I have been struggling with the next equality for some time now. The problem is, that I can get to the correct answers, but they are in the wrong intervals, and I can not figure out, what is it, that I am missing.

So:
| (3x-2)/(x-1) | = 2

I understand, that the root is 2/3, and that if x>= 2/3, the expression inside the absolute value sign is positive, and negative otherwise. After taking the different cases into consideration, I get that if x is the element of (-inf,2/3), the solution is 4/5, and if x is the element of [2/3,inf), the solution is 0. I simply don't know, what am I doing wrong, I have had no problems, solving absolute value equations before, but this rational one confuses me a bit.
Clearly, (though maybe not) this is an inequality, since if it is an equality, the solutions are 0, 4/5.

Assuming this is $|\frac{3x-2}{x-1}| > 2$

you require the fraction to be >2 or <-2

$\frac{3x-2}{x-1}>2$ when we have +/+ or -/-

$\frac{3x-2}{x-1}<-2$ when we have +/- or -/+

x>1 and 3x>2 for +/+

$3x-2>2x-2\Rightarrow\ x>0\Rightarrow\ x>1$

x<1 and 3x<2 for -/-

$3x-2<2x-2\Rightarrow\ x<0$

x>1 and 3x<2 for -/+

$3x-2<-2(x-1)\Rightarrow\ 5x<4$ contradiction as x>1

x<1 and 3x>2 for +/-

$3x-2>-2(x-1)\Rightarrow\ 5x>4\Rightarrow\ 0.8

All in all, you have x>0.8 and x<0

(sorry, my modem was jammed and saw no responses)

7. Thank you, it worked out flawlessly!

8. Thank you, though it was in fact an equality, I just had some (embarrasing) mehtodical problems. Thanks for your reply though, I will check the problem as an enaquality and it might come in handy.

9. Hello, szucslaszlo!

. . $\displaystyle \left|\frac{3x-2}{x-1}\right| \:=\: 2$

I blush to confess that I got tangled up in my algebra,
. . so I tried a graphical approach.

The graph of $y \:=\:\dfrac{3x-2}{x-1}$ looks like this:

Code:

|
|         :
|         :*
|         :
|         : *
|         :  *
|         :    *
|         :       *
3|         :             *
- - - + - - - - : - - - - - - - -
*   |         :
*         :
|  *      :
--------+----*----:------------------
|      *  :1
|         :
|       * :
|         :
|         :
|        *:
|         :
|         :
|         :
|        *:
|

In the graph of $y \,=\,\left|\dfrac{3x-2}{x-1}\right|$
. . everything below the $x$-axis is reflected upward.

Code:

|
|         :
|         :*
|        *:
|         : *
|         :  *
|         :    *
|        *:       *
3|         :             *
- - - + - - - - : - - - - - - - -
*   |       * :
*         :
|  *   *  :
--------+----*----:------------------
|         :1
|         :
|

And we want the points where $y \,=\,2.$

Code:

|
|         :
|         :*
|         :
|         : *
|         :  *
|         :    *
|        *:       *
|         :             *
. . . + . . . . : . . . . . . . .
*   |        *:
- - - ♥ - - - ♥ : - - - - -  y = 2
|  *   *  :
-------+----*----:------------------
|         :1
|         :
|

Now that I can see the solution, I can do the algebra.

The answers are: . $x \:=\:0,\,\frac{4}{5}$

,
,

,

,

,

,

,

,

,

,

,

,

,

,

Rational ineqyalities of modulus

Click on a term to search for related topics.