Reciprocal Function help

[GenericR]

First of all, I'm not exactly sure if this is the correct place to post this but I think it is.

I'm having a small problem with 2 questions, I've looked back into what notes I have and different online pdf's still can't find anything. I'm sure its a super easy solution but I just can't seem to recall it.

1. Polluted water from a factory flows into a pond. The concentration of pollutant (kg/m³), c, in the pond at time t minutes is modeled by the equation:

c(t)= 8 - 8000[1/(1000+2t)]

When will the concentration of pollutant in the pond reach 6 kg/m³?

Here's what I've got:

8 - 8000[1/(1000+2t)] = 6
2 - 8000[1/(1000+2t)] = 0
2 - [8000/8000000 + 16000t] = 0
{2(8000000 + 16000t) -8000}/[8000000 + 16000t] = 0
16000000 + 32000t - 8000 = 0

But that gives me a negative number so i'm confused, any help would be appreciated.

2. Solve [(2x - 1)/(x - 5)] > [(x + 1)/(x+5)]

I'm 95% sure I got this one but if anyone wants to give me a once over just in words then that would be great aswell.

Thanks for your time and help

 

[soroban]

Hello, GenericR!

1. Polluted water from a factory flows into a pond.
The concentration of pollutant (kg/m³), c, in the pond at time t minutes is modeled by the equation:
. . $\displaystyle c(t)\:=\:8 - 8000\left(\dfrac{1}{1000+2t}\right)$

When will the concentration of pollutant in the pond reach 6 kg/m³ ?

We have: . . . .. $\displaystyle c \:=\:8 - \dfrac{8000}{1000+2t}$

If $\displaystyle c = 6\!:$ . . . . . $\displaystyle 6 \:=\:8 - \dfrac{8000}{1000+2t}$

Subtract 8: . . $\displaystyle -2 \;=\;-\dfrac{8000}{1000+2t}$

Divide by -2: . . $\displaystyle 1 \;=\;\dfrac{4000}{1000+2t}$

Then: .$\displaystyle 1000 + 2t \;=\;4000$

Can you finish?

$\displaystyle \text{2. Solve: }\:\dfrac{2x - 1}{x - 5} \:>\: \dfrac{x + 1}{x+5}$

Why didn't you show us your answer?

I might have answered you with one word: Right!

Instead you're forcing me to solve the problem
. . and show you my answer.

 

[GenericR]

Why didn't you show us your answer?

I might have answered you with one word: Right!

Instead you're forcing me to solve the problem
. . and show you my answer.

That seems fair enough, here is my answer:

0 > [(x+1)/(x+5)] - [(2x-1)/(x-5)]
0 > [(x+5)(x-5) - (2x-1)(x+5)] / [(x+5)(x-5)]
0 > [x² - 4x - 5 - (2x² + 9x - 5)] / [(x+5)(x-5)]
0 > (-x² - 13x) / [(x+5)(x-5)]

Interval | -∞ < x < -5 | -5 < x < 5 | 5 < x < ∞ |
Value | -6 | 1 | 6 |
Sign(Above or below 0) | + | + | - |


Therefore, [(x+1)/(x+5)] - [(2x-1)/(x-5)] is less than 0 when (5 < x < ∞).

Thanks again for all the help!

 

[HallsofIvy]

That seems fair enough, here is my answer:

0 > [(x+1)/(x+5)] - [(2x-1)/(x-5)]
0 > [(x+5)(x-5) - (2x-1)(x+5)] / [(x+5)(x-5)]

Your first term should be (x+1)(x- 5) but I suspect that was a typo, as you have the correct expression below.

0 > [x² - 4x - 5 - (2x² + 9x - 5)] / [(x+5)(x-5)]
0 > (-x² - 13x) / [(x+5)(x-5)]

This not defined at x=-5 and x= 5 but is 0 at x= 0 and x= -13 so can change sign there.

Interval | -∞ < x < -5 | -5 < x < 5 | 5 < x < ∞ |
Value | -6 | 1 | 6 |
Sign(Above or below 0) | + | + | - |

You did not look at x< -13, -5< x< 0 or 0< x< 5

Therefore, [(x+1)/(x+5)] - [(2x-1)/(x-5)] is less than 0 when (5 < x < ∞).

Thanks again for all the help!

 

[Deleted member 4993]

First of all, I'm not exactly sure if this is the correct place to post this but I think it is.

I'm having a small problem with 2 questions, I've looked back into what notes I have and different online pdf's still can't find anything. I'm sure its a super easy solution but I just can't seem to recall it.

1. Polluted water from a factory flows into a pond. The concentration of pollutant (kg/m³), c, in the pond at time t minutes is modeled by the equation:

c(t)= 8 - 8000[1/(1000+2t)]

When will the concentration of pollutant in the pond reach 6 kg/m³?

Here's what I've got:

8 - 8000[1/(1000+2t)] = 6
2 - 8000[1/(1000+2t)] = 0
2 - [8000/8000000 + 16000t] = 0 ...... this is where it went wrong because $\displaystyle \ \ \ \displaystyle a * \frac{b}{c} \ = \ \frac{a*b}{c}$

so the line above should be:

2 - [8000/(1000+2t)] = 0


{2(8000000 + 16000t) -8000}/[8000000 + 16000t] = 0
16000000 + 32000t - 8000 = 0

But that gives me a negative number so i'm confused, any help would be appreciated.

2. Solve [(2x - 1)/(x - 5)] > [(x + 1)/(x+5)]

I'm 95% sure I got this one but if anyone wants to give me a once over just in words then that would be great as well.

Thanks for your time and help

.

 

[GenericR]

Your first term should be (x+1)(x- 5) but I suspect that was a typo, as you have the correct expression below.


This not defined at x=-5 and x= 5 but is 0 at x= 0 and x= -13 so can change sign there.


You did not look at x< -13, -5< x< 0 or 0< x< 5

I'm not sure what you mean, could you please clarify? Still a little confused on question #2.

Thanks for all the help!

 

[stapel]

I'm not sure what you mean, could you please clarify?

Which part are you finding confusing? What are your thoughts thus far? Where do you grind to a halt?

Please be complete. Thank you!

 

[GenericR]

Which part are you finding confusing? What are your thoughts thus far? Where do you grind to a halt?

Please be complete. Thank you!

Well I posted my work to this question above:

[FONT=MathJax_Main]2. Solve: [/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]5[/FONT]


0 > [(x+1)/(x+5)] - [(2x-1)/(x-5)]
0 > [(x+5)(x-5) - (2x-1)(x+5)] / [(x+5)(x-5)]
0 > [x² - 4x - 5 - (2x² + 9x - 5)] / [(x+5)(x-5)]
0 > (-x² - 13x) / [(x+5)(x-5)]

Interval | -∞ < x < -5 | -5 < x < 5 | 5 < x < ∞ |
Value | -6 | 1 | 6 |
Sign(Above or below 0) | + | + | - |


Therefore, [(x+1)/(x+5)] - [(2x-1)/(x-5)] is less than 0 when (5 < x < ∞).



Then HallsofIvy posted these corrections:

- This not defined at x=-5 and x= 5 but is 0 at x= 0 and x= -13 so can change sign there.
- You did not look at x< -13, -5< x< 0 or 0< x< 5

I'm not sure what I did wrong or how these corrections fit into my work, just overall confused.

Thanks again for all your help.

 

[JeffM]

Well I posted my work to this question above:

[FONT=MathJax_Main]2. Solve: [/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]>[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]5[/FONT]


0 > [(x+1)/(x+5)] - [(2x-1)/(x-5)]
0 > [(x+5)(x-5) - (2x-1)(x+5)] / [(x+5)(x-5)]
0 > [x² - 4x - 5 - (2x² + 9x - 5)] / [(x+5)(x-5)]
0 > (-x² - 13x) / [(x+5)(x-5)]

Interval | -∞ < x < -5 | -5 < x < 5 | 5 < x < ∞ |
Value | -6 | 1 | 6 |
Sign(Above or below 0) | + | + | - |


Therefore, [(x+1)/(x+5)] - [(2x-1)/(x-5)] is less than 0 when (5 < x < ∞).



Then HallsofIvy posted these corrections:

- This not defined at x=-5 and x= 5 but is 0 at x= 0 and x= -13 so can change sign there.
- You did not look at x< -13, -5< x< 0 or 0< x< 5

I'm not sure what I did wrong or how these corrections fit into my work, just overall confused.

Thanks again for all your help.

In most respects, solving inequations is no different from solving equations. But it is important to be aware of the differences.

The most important difference involves multiplication (and division) by a negative number..

$\displaystyle a > b\ and\ c < 0 \implies ac < bc,\ but\ a = b\ and\ c < 0 \implies ac = bc.$ The direction of inequality changes.

$\displaystyle a < b\ and\ c < 0 \implies ac > bc,\ but\ a = b\ and\ c < 0 \implies ac = bc.$ Again, the direction of inequality changes.

Furthermore, multiplication by zero is another difference.

$\displaystyle a > b\ and\ c = 0 \implies ac = bc,\ but\ a = b\ and\ c = 0 \implies ac = bc.$ The inequation becomes an equation.

$\displaystyle a < b\ and\ c < 0 \implies ac > bc,\ but\ a = b\ and\ c < 0 \implies ac = bc.$ Again, the inequation becomes an equation.

What Halls of Ivy was doing was to indicate that whenever you have inequations involving multiplication or division, you probably must consider multiple cases when signs change.

Do you follow all that? It is really important that you do, or inequations will drive you nuts.

Another thing to remember about inequations is that the solution is one or more ranges of numbers, not a unique number.

So what Halls of Ivy was doing was telling you that the range or ranges that solve the inequation are critically involved with where the function changes sign.

Now $\displaystyle \dfrac{2x - 1}{x - 5}$ is not even defined at x = 5 so clearly 5 is not in the range where the inequation is true.

And $\displaystyle \dfrac{x + 1}{x + 5}$ is not defined at x = - 5 so - 5 is not in the range where the inequation is true.

To solve this inequation, you need to multiply by (x - 5) and (x + 5). But multiplying by 0 changes the inequation into an equation and multiplying by a negative reverses the direction of inequality.

So Halls of Ivy gave you the clues to where you need to analyze to determine the range or ranges where the inequation is true. It is not true everywhere.

So give it another go, and show us your work so we can make sure you got the correct answer. If you are not sure, don't worry, and show your work anyway: we bark but never bite.

 

[GenericR]

So give it another go, and show us your work so we can make sure you got the correct answer. If you are not sure, don't worry, and show your work anyway: we bark but never bite.

Okay! Here's another try:

[FONT=Verdana, Arial, Tahoma, Calibri, Geneva, sans-serif]Solve[(2x - 1)/(x - 5)] > [(x + 1)/(x+5)][/FONT]

[FONT=Tahoma, Calibri, Verdana, Geneva, sans-serif]0[/FONT]>[(x+1)/(x-5)] - [(2x-1)/(x-5)]
[FONT=Tahoma, Calibri, Verdana, Geneva, sans-serif]0> [(x+5)(x-5) - (2x-1)(x+5)] / [(x+5)(x-5)][/FONT]
[FONT=Tahoma, Calibri, Verdana, Geneva, sans-serif]0> [x² - 4x - 5 - (2x² + 9x - 5)] / [(x+5)(x-5)][/FONT]
[FONT=Tahoma, Calibri, Verdana, Geneva, sans-serif]0> (-x² - 13x) / [(x+5)(x-5)][/FONT]

[FONT=Tahoma, Calibri, Verdana, Geneva, sans-serif]Interval| x[/FONT] <-13 | -13 < x < -5 | -5 < x < 0 | 0 < x < 5 | x <∞|
[FONT=Tahoma, Calibri, Verdana, Geneva, sans-serif]Value| -14 | -6 | -2 | 1 | 2 | 6 |[/FONT]
[FONT=Tahoma, Calibri, Verdana, Geneva, sans-serif]Sign(Aboveor below 0) | + | + | - | + | + | - |[/FONT]


[FONT=Tahoma, Calibri, Verdana, Geneva, sans-serif]Therefore,[[/FONT](x+1)/(x+5)]- [(2x-1)/(x-5)] is less than 0 when (-5 < x < 0) and (x <∞).

Am I correct in where I took hallsofivy's help?

Thanks again!

 

[JeffM]

Okay! Here's another try:

Solve[(2x - 1)/(x - 5)] > [(x + 1)/(x+5)]

0>[(x+1)/(x-5)] - [(2x-1)/(x-5)]
0> [(x+5)(x-5) - (2x-1)(x+5)] / [(x+5)(x-5)]
0> [x² - 4x - 5 - (2x² + 9x - 5)] / [(x+5)(x-5)]
0> (-x² - 13x) / [(x+5)(x-5)]

Interval| x<-13 | -13 < x < -5 | -5 < x < 0 | 0 < x < 5 | x <∞|
Value| -14 | -6 | -2 | 1 | 2 | 6 |
Sign(Aboveor below 0) | + | + | - | + | + | - |


Therefore,[(x+1)/(x+5)]- [(2x-1)/(x-5)] is less than 0 when (-5 < x < 0) and (x <∞).

Am I correct in where I took hallsofivy's help?

Thanks again!

Here is the way that I would do this problem.

First, note that x = 5 and x = - 5 are excluded from the range because either case would involve division by zero. Those values of x are not permissible.

Second, reverse what is to be shown.

$\displaystyle \dfrac{2x - 1}{x - 5} \le \dfrac{x + 1}{x + 5}.$

If I exclude any permissible x that satisfies the inequation above, I have identified the values of x that satisfy $\displaystyle \dfrac{2x - 1}{x - 5} > \dfrac{x + 1}{x + 5}.$

This kind of backwards logic is called modus tollens.

To get rid of the fractions I need to multiply by (x + 5)(x - 5). If x is neither 5 nor - 5, then (x + 5)(x - 5) > 0 or
(x - 5)(x + 5) < 0.

If (x - 5)(x + 5) > 0, then either (x - 5) < 0 and (x + 5) < 0 or (x - 5) > 0 and (x + 5) > 0. Let those be cases I and II respectively.

If (x - 5)(x + 5) < 0, then (x - 5) < 0 and (x + 5) > 0. Let that be case III.

Case I: (x - 5) < 0 and (x + 5) < 0 so x < - 5 and (x - 5)(x + 5) > 0.

$\displaystyle \dfrac{2x - 1}{x - 5} \le \dfrac{x + 1}{x + 5} \implies (x - 5)(x + 5) * \dfrac{2x - 1}{x - 5} \le (x - 5)(x + 5) * \dfrac{x + 1}{x + 5} \implies (x + 5)(2x - 1) \le (x - 5)(x + 1) \implies$

$\displaystyle 2x^2 + 9x - 5 \le x^2 - 4x - 5 \implies x^2 + 13x \le 0 \implies x(x + 13) \le 0 \implies \dfrac{x(x + 13)}{x} \ge \dfrac{0}{x} \implies x \ge -13.$

But this tells us what range we do not want. So x < - 5 and x < - 13 or x < - 13.

Case II: (x - 5) > 0 and (x + 5) > 0 so x > 5 and (x - 5)(x + 5) > 0.

$\displaystyle \dfrac{2x - 1}{x - 5} \le \dfrac{x + 1}{x + 5} \implies (x - 5)(x + 5) * \dfrac{2x - 1}{x - 5} \le (x - 5)(x + 5) * \dfrac{x + 1}{x + 5} \implies (x + 5)(2x - 1) \le (x - 5)(x + 1) \implies$

$\displaystyle 2x^2 + 9x - 5 \le x^2 - 4x - 5 \implies x^2 + 13x \le 0 \implies x(x + 13) \le 0 \implies \dfrac{x(x + 13)}{x} \le \dfrac{0}{x} \implies x \le -13.$

But this tells us what we do not want. So x > - 13 and x > 5 or x > 5.

Case III: (x - 5) < 0 and (x + 5) > 0 so - 5 < x < 5 and (x - 5)(x + 5) < 0.

$\displaystyle \dfrac{2x - 1}{x - 5} \le \dfrac{x + 1}{x + 5} \implies (x - 5)(x + 5) * \dfrac{2x - 1}{x - 5} \ge (x - 5)(x + 5) * \dfrac{x + 1}{x + 5} \implies (x + 5)(2x - 1) \ge (x - 5)(x + 1) \implies$

$\displaystyle 2x^2 + 9x - 5 \ge x^2 - 4x - 5 \implies x^2 + 13x \ge 0 \implies x(x + 13) \ge 0.$

That is true if x = 0. And it is true if x > 0. But this tells us what we do not want. So - 5 < x < 0.

Thus we have three ranges x < -13, -5 < x < 0, and 5 < x.

Let's check.

x = - 15.

$\displaystyle \dfrac{2 * (-15) - 1}{-15 - 5} = \dfrac{-31}{-20} = \dfrac{31}{20}.$

$\displaystyle \dfrac{-15 + 1}{-15 + 5} = \dfrac{-14}{-10} = \dfrac{28}{20} < \dfrac{31}{20}.$ It checks.

x = - 1.

$\displaystyle \dfrac{2(-1) - 1}{-1 - 5} =\dfrac{-3}{-6} = \dfrac{1}{2}.$

$\displaystyle \dfrac{-1 + 1}{-1 + 5} = \dfrac{0}{4} = 0 < \dfrac{1}{2}.$ It checks.

x = 6

$\displaystyle \dfrac{2 * 6 - 1}{6 - 5} = \dfrac{11}{1} = 11.$

$\displaystyle \dfrac{6 + 1}{6 + 5} = \dfrac{7}{11} < 1 < 11.$ It checks.

 

[GenericR]

Here is the way that I would do this problem.

First, note that x = 5 and x = - 5 are excluded from the range because either case would involve division by zero. Those values of x are not permissible.

Second, reverse what is to be shown.

$\displaystyle \dfrac{2x - 1}{x - 5} \le \dfrac{x + 1}{x + 5}.$

If I exclude any permissible x that satisfies the inequation above, I have identified the values of x that satisfy $\displaystyle \dfrac{2x - 1}{x - 5} > \dfrac{x + 1}{x + 5}.$

This kind of backwards logic is called modus tollens.

To get rid of the fractions I need to multiply by (x + 5)(x - 5). If x is neither 5 nor - 5, then (x + 5)(x - 5) > 0 or
(x - 5)(x + 5) < 0.

If (x - 5)(x + 5) > 0, then either (x - 5) < 0 and (x + 5) < 0 or (x - 5) > 0 and (x + 5) > 0. Let those be cases I and II respectively.

If (x - 5)(x + 5) < 0, then (x - 5) < 0 and (x + 5) > 0. Let that be case III.

Case I: (x - 5) < 0 and (x + 5) < 0 so x < - 5 and (x - 5)(x + 5) > 0.

$\displaystyle \dfrac{2x - 1}{x - 5} \le \dfrac{x + 1}{x + 5} \implies (x - 5)(x + 5) * \dfrac{2x - 1}{x - 5} \le (x - 5)(x + 5) * \dfrac{x + 1}{x + 5} \implies (x + 5)(2x - 1) \le (x - 5)(x + 1) \implies$

$\displaystyle 2x^2 + 9x - 5 \le x^2 - 4x - 5 \implies x^2 + 13x \le 0 \implies x(x + 13) \le 0 \implies \dfrac{x(x + 13)}{x} \ge \dfrac{0}{x} \implies x \ge -13.$

But this tells us what range we do not want. So x < - 5 and x < - 13 or x < - 13.

Case II: (x - 5) > 0 and (x + 5) > 0 so x > 5 and (x - 5)(x + 5) > 0.

$\displaystyle \dfrac{2x - 1}{x - 5} \le \dfrac{x + 1}{x + 5} \implies (x - 5)(x + 5) * \dfrac{2x - 1}{x - 5} \le (x - 5)(x + 5) * \dfrac{x + 1}{x + 5} \implies (x + 5)(2x - 1) \le (x - 5)(x + 1) \implies$

$\displaystyle 2x^2 + 9x - 5 \le x^2 - 4x - 5 \implies x^2 + 13x \le 0 \implies x(x + 13) \le 0 \implies \dfrac{x(x + 13)}{x} \le \dfrac{0}{x} \implies x \le -13.$

But this tells us what we do not want. So x > - 13 and x > 5 or x > 5.

Case III: (x - 5) < 0 and (x + 5) > 0 so - 5 < x < 5 and (x - 5)(x + 5) < 0.

$\displaystyle \dfrac{2x - 1}{x - 5} \le \dfrac{x + 1}{x + 5} \implies (x - 5)(x + 5) * \dfrac{2x - 1}{x - 5} \ge (x - 5)(x + 5) * \dfrac{x + 1}{x + 5} \implies (x + 5)(2x - 1) \ge (x - 5)(x + 1) \implies$

$\displaystyle 2x^2 + 9x - 5 \ge x^2 - 4x - 5 \implies x^2 + 13x \ge 0 \implies x(x + 13) \ge 0.$

That is true if x = 0. And it is true if x > 0. But this tells us what we do not want. So - 5 < x < 0.

Thus we have three ranges x < -13, -5 < x < 0, and 5 < x.

Let's check.

x = - 15.

$\displaystyle \dfrac{2 * (-15) - 1}{-15 - 5} = \dfrac{-31}{-20} = \dfrac{31}{20}.$

$\displaystyle \dfrac{-15 + 1}{-15 + 5} = \dfrac{-14}{-10} = \dfrac{28}{20} < \dfrac{31}{20}.$ It checks.

x = - 1.

$\displaystyle \dfrac{2(-1) - 1}{-1 - 5} =\dfrac{-3}{-6} = \dfrac{1}{2}.$

$\displaystyle \dfrac{-1 + 1}{-1 + 5} = \dfrac{0}{4} = 0 < \dfrac{1}{2}.$ It checks.

x = 6

$\displaystyle \dfrac{2 * 6 - 1}{6 - 5} = \dfrac{11}{1} = 11.$

$\displaystyle \dfrac{6 + 1}{6 + 5} = \dfrac{7}{11} < 1 < 11.$ It checks.

I'm sorry and I don't mean to sound rude but I couldn't follow anything you said. Well not anything I understand the checks and the idea of different "cases" but I can't seem to follow your work and get your answers, It's okay though I've already handed in my work for this question. I will ask my teacher for help when I get the work back. Thanks again for all of your help everyone!

 

[Deleted member 4993]

Solve [(2x - 1)/(x - 5)] > [(x + 1)/(x+5)]

Jeff has already given you the "right" way to solve this problem. The way I would do this would be same - jumping some steps.

Here we need to multiply both sides by (x-5)*(x+5) - watching the sign because if we multiply by negative number - the inequality sign flip-flops.

So

(x-5)(x+5)>0 when x>5 OR x<-5 (|x| >5)

(x-5)(x+5)<0 when -5<x<5

so when x>5 OR x<-5

[(2x - 1)/(x - 5)] > [(x + 1)/(x+5)] → [(2x - 1)/(x - 5)] (x-5)(x+5)> [(x + 1)/(x+5)](x-5)(x+5) ..... no flip flop of .> because (x-5)(x+5)>0

(2x-1)(x+5) > (x+1)(x-5) → 2x² + 9x - 5 > x² - 4x - 5 → x² + 13x > 0 → x(x-13) > 0 → x >13 or x<0 → But |x|>5 so x>13 and x<-5

Now when -5< x <5 → |x| < 5

[(2x - 1)/(x - 5)] > [(x + 1)/(x+5)] → [(2x - 1)/(x - 5)] (x-5)(x+5)< [(x + 1)/(x+5)](x-5)(x+5) ..... flip flop of .> because (x-5)(x+5)<0

[(2x - 1)/(x - 5)] (x-5)(x+5)< [(x + 1)/(x+5)](x-5)(x+5) → x(x-13) < 0 → 0 < x <13 or → But |x|<5 so 0<x<5

So

[(2x - 1)/(x - 5)] > [(x + 1)/(x+5)] for x < -5 or 0< x<5 or x>13 (three distinct region where the proposed inequality is true)

That also means

[(2x - 1)/(x - 5)] ≤ [(x + 1)/(x+5)] for -5 < x ≤0 or 5 < x ≤ 13

Notice that x = ± 5 (running out of red ink) is excluded because the expression [(2x - 1)/(x - 5)] - [(x + 1)/(x+5)] is not defined at x = ±5

go to

http://www.wolframalpha.com/i

and confirm your finding.