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1/22/2020 12:12:56 PM
Posted: 9/17/2009 4:52:30 PM EST
Ok, this is really stupid, but we can't figure this one out.

We need to solve this inequality.

3/(x-1) - 4/x >= 1

So x =/= 1 or 0

We bring the 1 over so we have

3/(x-1) - 4/x - 1 >= 0

Then get a common denominator and simplify

(x^2 -2x +4)/(x(x-1)) >= 0

What now? That numerator doesn't factor
Link Posted: 9/17/2009 4:59:15 PM EST
[Last Edit: 9/17/2009 5:04:28 PM EST by wheel]
The key is to get rid of the fractions by multiplying everything by a common denominator, in this case x(x-1):

3x(x-1)/(x-1) - 4x(x-1)/x >= x(x-1)

3x -4(x-1) >= x(x-1)

3x -4x + 4 >= x^2 - x

4 >= x^2

x <= 2

Link Posted: 9/17/2009 5:06:32 PM EST
Originally Posted By wheel:
The key is to get rid of the fractions by multiplying everything by a common denominator, in this case x(x-1):

3x(x-1)/(x-1) - 4x(x-1)/x >= x(x-1)

3x -4(x-1) >= x(x-1)

3x -4x + 4 >= x^2 - x

4 >= x^2

x <= 2



Maybe, I dunno. Our Professor made a whole big deal about not multiplying by the denominator today. Can't quite remember why though.
Link Posted: 9/17/2009 5:10:24 PM EST
[Last Edit: 9/17/2009 5:15:00 PM EST by wheel]
Uhh..... it works. Try plugging in 2 for x, then try 1.9 for instance.


ETA: actually I goofed on the last line:

x^2 <= 4

-2 >= x <= 2
Link Posted: 9/17/2009 5:20:06 PM EST
[Last Edit: 9/17/2009 5:26:56 PM EST by Ryan1021]
Edit again:

This is what I got

-2<=x<0
1<x<=2

Link Posted: 9/17/2009 5:42:27 PM EST
[Last Edit: 9/17/2009 5:43:37 PM EST by nicholsmf]
Simple?

1 + X = 2

is simple

3/(x-1) - 4/x >= 1

So x =/= 1 or 0

is not!
Link Posted: 9/17/2009 5:43:58 PM EST
[Last Edit: 9/17/2009 5:45:42 PM EST by msprooch]
Originally Posted By Feier:
Originally Posted By wheel:
The key is to get rid of the fractions by multiplying everything by a common denominator, in this case x(x-1):

3x(x-1)/(x-1) - 4x(x-1)/x >= x(x-1)

3x -4(x-1) >= x(x-1)

3x -4x + 4 >= x^2 - x

4 >= x^2

x <= 2



Maybe, I dunno. Our Professor made a whole big deal about not multiplying by the denominator today. Can't quite remember why though.


Hes an idiot.

EDIT the only time you have to worry about multipling or dividing on an inequallity is when you are x or / by a negative number.
Link Posted: 9/17/2009 5:46:55 PM EST
Originally Posted By nicholsmf:
Simple?

1 + X = 2

is simple

3/(x-1) - 4/x >= 1

So x =/= 1 or 0

is not!


compared to a triple integral, its is simple
Link Posted: 9/17/2009 5:51:39 PM EST
Originally Posted By nicholsmf:
Simple?

1 + X = 2

is simple

3/(x-1) - 4/x >= 1

So x =/= 1 or 0

is not!


Not dividing by 0 IS simple.
Link Posted: 9/17/2009 6:46:18 PM EST
[Last Edit: 9/17/2009 6:56:57 PM EST by Mudfish]
1) Bring the 1 over: 3/(x-1) - 4/x - 1 >= 0.

2) Multiply by common denominator: 3x - 4(x-1) / x(x-1) - x(x-1) / x(x-1) >= 0.

3) Combine: 3x - 4x + 4 -x^2 + x / x(x-1) >= 0.

4) Combine: 4 - x^2 / x(x-1) >= 0.

5) Factor numerator: (2+x) (2-x) / x(x-1) >= 0. For simplicity, we will call the left side of this inequality f(x).

6) Note: f(x) can change signs only when there is a zero in the numerator or denominator.
i.e. x = -2, 0, 1, 2. So we must check the intervals on both sides of these critical points.

7) Check interval x < -2: (plug in x = -3), f(x) < 0, so this is NOT part of the solution set for f(x) >= 0.

8) Check interval -2 < x < 0 (plug in x = -1), f(x) > 0, so this IS part of the solution set for f(x) >= 0.

9) Check interval 0 < x < 1 (plug in x = 1/2), f(x) < 0, so this is NOT part of the solution set for f(x) >= 0.

10) Check interval 1 < x < 2 (plug in x = 3/2), f(x) > 0, so this IS part of the solution set for f(x) >= 0.

11) Check interval x > 2 (plug in x = 3), f(x) < 0, so this is NOT part of the solution set for f(x) >= 0.

So our solution set is: { -2 <= x < 0, 1 < x <= 2 }. (Note that f(x) is undefined for x = 0 and x = 1).










Link Posted: 9/17/2009 7:28:41 PM EST
[Last Edit: 9/17/2009 7:31:23 PM EST by Meche_03]
I believe the MathFu in this post is weak...........

your teacher is correct as far as I know, you can not multiply by the denominator when working with INEQUALITIES

You have to get each section to have the same denominator though .. x(x-1)

first subtract one from both sides then multiply each section by a unit of one be it x/x, x-1/x-1, x(x-1)/x(x-1)

3x 4(x-1) x(x-1) >= 0
––––- - ––––––––- - ––––––-
x(x-1) x(x-1) x(x-1)

then combine all over one denominator

3x - 4(x-1) - x(x-1) >= 0
––––––––––––––––––––––––-
x(x-1)

simplifiy

3x - 4x +4 - x^2 + x >= 0
––––––––––––––––––––––-
x(x-1)

simplify

-x^2 + 4 >= 0
––––––––––-
x(x-1)

factor

(x-2)(-x-2) >= 0
––––––––––––––
x(x-1)

solve each factor equal to zero, this gives your points where the line crosses zero and changes signs

x= 2, -2, 0, 1

plot matrix for sign of each factor when between each point, then at the bottom of matrix multiply signs, ie -*-=+, -*+=-, +*+=+

-2 0 1 2
x - - + + +
-x-2 + - - - -
x-2 - - - - +
x-1 - - - + +
––––––––––––––––––––––––––––––––-
- + - + -

only the region with positives are greater than zero as required above with the equation >= 0

thus -2 <= x < 0 and 1 < x <= 2 , x can not be = to 0 or 1 since they are the cross points found in the denominator

[-2,0),(1,2]

I think

F* math equations on forums, the format gets screwed up

damn, seconds to late
Mudfish's MathFu is strong
Link Posted: 9/17/2009 7:56:54 PM EST
Originally Posted By nicholsmf:
Simple?

1 + X = 2

is simple

3/(x-1) - 4/x >= 1

So x =/= 1 or 0

is not!


It's damn simple compared to Triple Integrals, Complex Partial Differential Equations, and Fourier Series.
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