Solving an Inequation with Fractions: 1° Equation Help

[Brunno]

Hi friends!
Could you help me solve this inequation.I tried and tried but i could only get to 3° equation and the fact is that this should result in a 1° equation.So...

\frac{1}{x-1} +\frac{2}{x-2} - \frac{3}{x-3} < 0

 

[arildno]

Well, what have you done?

Add the fractions together, and much will simplify!

 

[Brunno]

Ok,I will try that again and then i post here any doubt.

 

[Brunno]

Is this the right way to the solution?

\frac{2(x-1)+(x-2)+3(x-1)(x-2)}{(x-1)(x-2)(x-3)}

 

[Mark44]

Is this the right way to the solution?

\frac{2(x-1)+(x-2)+3(x-1)(x-2)}{(x-1)(x-2)(x-3)}

No. You need to multiply the first term in the inequality (there's no such word as inequation) by (x - 2)(x - 3) over itself, multiply the second term by (x - 1)(x - 3) over itself, and multiply the third term by (x - 1)(x - 2) over itself.

You sort of did this for the third term, but didn't notice that the numerator was -3, not +3. I can't tell what you did for the first and second terms.

 

[Brunno]

Then i simplify and get to:

\frac{6x^2-4x+6}{x^3-6x^2+11x-6}

 

[Mark44]

No, there is no x² term in the numerator. Also, why did you expand the denominator? You really don't want to do that.

Also, let's not lose sight of the fact that you're working with an inequality. Add the < 0 back in on the right side.

 

[Brunno]

\frac{(x-2)(x-3)+2(x-1)(x-3)-3(x+2)(x-1)}{(x-1)(x-2)(x-3)}

So if there's no X² on the numerator i don't know how to go any futher.

 

[Mentallic]

Look at the numerator, the x² terms from the first set of factors will be x², then in the next it will be 2x², then -3x² which obviously cancel to 0.

Show us how you expanded.

 

[Brunno]

Then I get to:


\frac{-16x+18}{(x-1)(x-2)(x-3)}

 

[Mark44]

No, both terms in the numerator are incorrect. After combining the three fractions the numerator should be (x - 2)(x - 3) + 2(x - 1)(x - 3) - 3(x - 1)(x - 2). Starting here, show how you are simplifying the numerator. Don't show us just your final expression - show the intermediate work as well.

 

[Brunno]

This was what i did:

(x-2)(x-3) = x²-3x-2x+6=x²-5x+6
(2x-2)(x-3)= 2x²-6x-2x+6 =2x²-8x+6
(-3x-6)(x-1) =-3x²+3x-6x+6

adding all together:-16x+18

 

[HallsofIvy]

This was what i did:

(x-2)(x-3) = x²-3x-2x+6=x²-5x+6
(2x-2)(x-3)= 2x²-6x-2x+6 =2x²-8x+6
(-3x-6)(x-1) =-3x²+3x-6x+6

Here is your error: -3(x- 2)= -3x+ 6, not -3x- 6.

adding all together:-16x+18

 

[Brunno]

\frac{-12x+12}{(x-1)(x-2)(x-3)}= \frac{12}{(x-2)(x-3)}?

 

[Brunno]

\frac{-12x+12}{(x-1)(x-2)(x-3)}= \frac{-12}{(x-2)(x-3)}?

 

[Mark44]

\frac{-12x+12}{(x-1)(x-2)(x-3)}= \frac{12}{(x-2)(x-3)}?

No.

\frac{-12x+12}{(x-1)(x-2)(x-3)}= \frac{-12}{(x-2)(x-3)}?

No.

 

[Mark44]

This was what i did:

(x-2)(x-3) = x²-3x-2x+6=x²-5x+6
(2x-2)(x-3)= 2x²-6x-2x+6 =2x²-8x+6
(-3x-6)(x-1) =-3x²+3x-6x+6

adding all together:-16x+18

The first two equations are OK. The last is not. Simplify -3(x - 2)(x - 1). To make things easier on yourself, multiply x - 2 and x - 1 together first, then multiply by -3.

 

[Brunno]

So,for now I'm done.I have a lot of stuff to learn and this is really pissing me off.I intend to passe on a kind of brazillian SAT and this is a question i found in a chapter about 1o degree functions and could not solve.But it is just a detail to me.But i really would love to see how to solve it.But anyways,it's just one more problem and i have a lot more complicated.Hope you guys understand!
Thank you guys very much!

 

[arildno]

So,for now I'm done.I have a lot of stuff to learn and this is really pissing me off.I intend to passe on a kind of brazillian SAT and this is a question i found in a chapter about 1o degree functions and could not solve.But it is just a detail to me.But i really would love to see how to solve it.But anyways,it's just one more problem and i have a lot more complicated.Hope you guys understand!
Thank you guys very much!

What sort of attitude is this??

You show deep problems with handling elementary algebra, and this "pisses you off"??

Unless you hone your skills in this, you will necessarily stagnate in your mathematical development.

Is that what you really want?

 

[Mark44]

I agree with arildno. You will not be able to do the more complicated problems if you are unable to carry out elementary algebra operations such as the ones in this problem.

 

[Brunno]

So why don't you guys just put the steps so i can see how i can reach a solution.I REALLY want to understand this prpblem!

Thank you!

 

[Mark44]

The best way for you to learn something is for you to do the work, because then it sticks in your mind. If we were to do the work by putting in all the steps, you might be able to understand it, but you probably wouldn't be able to put the steps into practice on a similar problem.

Besides that, the rules of this forum do not permit us to show complete solutions to problems.

 

[Brunno]

So it means you guys can't help me more then this?Sad to read this,I really don't believe it.I know much people here who helps knows that showing one a solution of a problem may help a lot.What really do teacher do in schools?They show a problem and when the pupil don't know how to get to the solution the professor shows how.That's not happening in this thread.

Thank you!

 

[diazona]

What really do teacher do in schools?They show a problem and when the pupil don't know how to get to the solution the professor shows how.

No they don't. (Well some do, but their students do not learn very much.)

Anyway, for what it's worth, you were getting close.

 

[Brunno]

Well,this is what you guys think and i respect that but each one is different.Not everything is mathematics,there's no a formula for each student that fix very well for all of us.So for me a resolution or at least the steps to it would be a great deal.what may not be for others students as you say.

Thank you.

 

[Mark44]

So it means you guys can't help me more then this?Sad to read this,I really don't believe it.

Here's from the Rules page, in the section titled Homework Help (https://www.physicsforums.com/showthread.php?t=414380). Emphasis added by me.

On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

I know much people here who helps knows that showing one a solution of a problem may help a lot.What really do teacher do in schools?They show a problem and when the pupil don't know how to get to the solution the professor shows how.That's not happening in this thread.

But we have pointed out exactly where your error is, so if you are as interested in seeing this problem through as you claim to be, go back to that step and fix what's wrong. As before, we'll be happy to check your work and steer you in the right direction, but we are NOT going to work the problem for you.

 

[Brunno]

Ok,thank you Mark and everybody.
I may anytime come back see the problem one more time and try to solve it again.

 

[arildno]

Well,this is what you guys think and i respect that but each one is different.Not everything is mathematics,there's no a formula for each student that fix very well for all of us.So for me a resolution or at least the steps to it would be a great deal.what may not be for others students as you say.

Here, you provide evidence for a fallacious self-understanding on your part.

The ONLY way to master elementary algebra, which you don't at the moment, is to DO the steps on your own.

It is because you show an irrational aversion of doing this that you have stagnated in mathematical development.

 

[Brunno]

Here, you provide evidence for a fallacious self-understanding on your part.

The ONLY way to master elementary algebra, which you don't at the moment, is to DO the steps on your own.

It is because you show an irrational aversion of doing this that you have stagnated in mathematical development.

But i already do it.Before even put this question here i tried the solution a thousand times.I guess now I'm just bored about doing it again.You know...

 

[Mark44]

I REALLY want to understand this prpblem!

... but not enough to actually do the work?

 

[Brunno]

I got to the solution!

\frac{1}{x-1}+\frac{2}{x-2}-\frac{3}{x-3}<0

\frac{1(x-2)(x-3)+2(x-1)(x-3)-3(x-1)(x-2)}{(x-1)(x-2)(x-3)}

(x-2)(x-3)=x²-3x-2x+6

(2x-2(x-3)=2x²-6x-2x+6=2x²-8x+6

-3x+3(x-2)=-3x²+6x+3x+6=-3x²+9x-6\frac{-4x+6}{(x-1)(x-2)(x-3)}

-4x+6<0 --> -4x<-6 --> x>3/2

x-1>0 --> x>1

x-2>0 --> x>2

x-3>0 --> x>3

S={x E R/x<1 or 3/2<x<2 or x>3}

So guys,I thank you all very much to show that i could do this by myself!

Now I believe "Yes we can".

 

[Mark44]

Good job!