Solving equations

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Probability

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2/x+1 + 2x - 3/x

I got to this stage;

2x^2 - x - 3 / x(x+1)

I asked myself would it factor?

(x + 1)(x - 3) = x^2 -3x + x - 3 = x^2 - 2x - 3 NO

2(x + 1)(x - 3) = 2(x^2 -2x - 3) = 2x^2 -4x - 6 NO

I also came up with 2x^2 - 2x - 3

I seem to be really struggling with trying to find a value for x?

any help much appreciated

Prob:confused:
 
Solving equations

2/x+1 + 2x - 3/x

Are you getting sloppy? You've posted 59 times already.

That's not an equation; all equations contain an equal sign.

Did you leave off your grouping symbols?

Your typing above means this:

\(\displaystyle \dfrac{2}{x} + 1 + 2x - \dfrac{3}{x}\)


:idea: Maybe other people have time for figuring out the presentation above; at the moment, I do not.

Please try to be complete. (Use the button for previewing your posts.)

Cheers
 
Last edited:
Yes it is an equation.

2 divided by (x + 1) + (2x - 3) divided by (x) = 0

Sorry for any confusion but not posted 59 times:D
 
Instead of combining the two algebraic ratios as your first step, multiply both sides of the given equation by the common denominator.

That will clear the fractions, leaving you with a quadratic equation in the form ax^2+bx+c=0.
 
2/x+1 + 2x - 3/x

I concur with mmm: you need to use grouping symbols carefully. We cannot see your book.

But you clarified below that you meant [2 / (x + 1)] + [(2x - 3) / x] = 0.

I got to this stage;

2x^2 - x - 3 / x(x+1) Here again you failed to use grouping symbols. You mean (2x^2 - x - 3) / [x(x + 1)]. However this is an error.

\(\displaystyle 0 = \dfrac{2}{x + 1} + \dfrac{2x - 3}{x} = \dfrac{2x}{(x + 1)x} + \dfrac{(2x - 3)(1 + x)}{x(x + 1)} = \dfrac{2x}{x(x + 1)} + \dfrac{2x + 2x^2 - 3 - 3x}{x(x + 1)} \implies\)

\(\displaystyle \dfrac{2x}{x(x + 1)} + \dfrac{2x^2 - x - 3}{x(x + 1} = \dfrac{2x^2 + x - 3}{x(x + 1)} = 0.\)

I asked myself would it factor?
(x + 1)(x - 3) = x^2 -3x + x - 3 = x^2 - 2x - 3 NO

2(x + 1)(x - 3) = 2(x^2 -2x - 3) = 2x^2 -4x - 6 NO

I also came up with 2x^2 - 2x - 3

I seem to be really struggling with trying to find a value for x?

any help much appreciated

Prob:confused:
.
Let's see you finish it up. Factoring is not the only method to solve a quadratic.
 
\(\displaystyle \dfrac{2}{x + 1} + \dfrac{2x - 3}{x} = 0 \implies\)


\(\displaystyle \dfrac{2x}{(x + 1)x} + \dfrac{(2x - 3)(x + 1)}{x(x + 1)} = 0 \implies\)


\(\displaystyle \dfrac{2x}{x(x + 1)} + \dfrac{2x^2 + 2x - 3x - 3}{x(x + 1)} = 0 \implies\)


\(\displaystyle \dfrac{2x}{x(x + 1)} + \dfrac{2x^2 - x - 3}{x(x + 1)} = 0 \implies \)


\(\displaystyle \dfrac{2x^2 + x - 3}{x(x + 1)} = 0 \)



Implications are between each equation.



And user Probability,

you cannot include any x-values for solutions that make the denominator equal zero.
 
you cannot include any x-values for solutions that make the denominator equal zero

Yes, with rational equations (especially when a denominator cancels during the solution process), it is crucial to check all of the resulting solutions to ensure that they work.
 
I must thnk you all for the time and effort you have put into replying, thank you for that, however, my maths program MATHCAD gives the following;

2 divided by (x + 1) + 2x - 3 divided by x implies 2(x^2 - x) - 3 divided by x(x + 1) implies 2(x^2 - x) - 3, which factors to;

2x^2 - 2x - 3

This seems slightly different to your answers and my tutors advise, however neither seem to work with the formula as I can't get 0 as an answer ?

Regards

Probability:confused:
 
my maths program MATHCAD gives the following:

2x^2 - 2x - 3

That is not correct. You are doing something wrongly.

:idea: The exercise is easy to solve using paper and pencil.

Multiply both sides of the given equation by the common denominator x(x+1).

The simplified result is: 2x^2 + x - 3 = 0
 
That is not correct. You are doing something wrongly.

:idea: The exercise is easy to solve using paper and pencil.

Multiply both sides of the given equation by the common denominator x(x+1).

The simplified result is: 2x^2 + x - 3 = 0

The equation says that the result = 0.

So I admit I am learning something new and latex is not in sight on this forum when entering results, so I can't practice it.

Given that 2x^2 + x - 3 = 0, using the quadratic formula I get the two results;

x = - 0.38, and

x = 0.6

If the equation does equal = 0, then I must be making a mistake somewhere that I can't see?
 
The equation says that the result = 0.

No. The equation says that the left-hand side is equal to zero.

The equation does not say anything about your result(s).


Given that 2x^2 + x - 3 = 0

No. It is given that 2/(x + 1) + (2x - 3)/x = 0

You rewrote the given.


using the quadratic formula I get the two results;

x = - 0.38, and

x = 0.6

I must be making a mistake somewhere that I can't see?

The word and should be or.

Yes, those solutions are incorrect. The correct solutions are:

x = -3/2 or x = 1

Nobody here will be able to see your mistakes until after you show your work.
 
No. The equation says that the left-hand side is equal to zero.

The equation does not say anything about your result(s).




No. It is given that 2/(x + 1) + (2x - 3)/x = 0

You rewrote the given.




The word and should be or.

Yes, those solutions are incorrect. The correct solutions are:

x = -3/2 or x = 1

Nobody here will be able to see your mistakes until after you show your work.

Yes my fault sorry, this is where I was getting it wrong in the final parts;

Using the quadratic formula I ended up with;

1 + or - square root 25 divided by 4. So far so good, then sticking to the rules of bidmas is where I went wrong, I was dividing 5 by 4 before adding or subtracting 1 from 5.

so 1 - 5 = 4 divided by 4 implies x = 1, or

1 + 5 = 6 divided by 4 implies x = 1.5.

Bidmas, correct me if I am wrong states Brackets (first), Powers (second), division (third), addition (fourth) and finally subtraction.

So addition and subtraction are first in this case followed by division.

If I am wrong please advise but if I am right this is where all my confusion in the latter part has occured, and even though I have completed many ways of doing this equation because the end result has always been wrong, I have contineously blamed the algrebra, so I have ended up very confused throughout.

I would however appreciate a little understanding from advanced members with less of the back stabbing, which is uncalled for, we all had a beginning with many misunderstandings at some point in time.

Thank you for all your help

Probability:smile:
 
Bidmas, correct me if I am wrong states Brackets (first), Powers (second), division (third), addition (fourth) and finally subtraction.

I'm not familiar with that acronym.

In the USA, we use this acronym: PEMDAS

The mnemonic is: "Please Excuse My Dear Aunt Sally".


P - parentheses (i.e., grouping symbols)

E - exponentiation

MD - multiplication/division

AS - addition/subtraction


I'm concerned that you wrote "and finally subtraction".

Addition and subtraction are done, in order as they appear, from left to right.

Multiplication and division are done, in order as they appear, from left to right.



Your solution has a sign error on x = 3/2.


\(\displaystyle \dfrac{-1 \pm \sqrt{1 + 24}}{2(2)}\)

\(\displaystyle \dfrac{-1 \pm 5}{4}\)

\(\displaystyle \dfrac{-1 + 5}{4} \text{OR} \dfrac{-1 - 5}{4}\)

\(\displaystyle \dfrac{4}{4} \text{OR} \dfrac{-6}{4}\)
 

It was not aimed at you, you have provided very good information and I thank you for that.

It was aimed here;

NOTE: if it was up to me, I'd cause your posts to vanish if you
don't learn to post PROPERLY.

Some people as good as they are forget that some others are learning the ropes.

Thanks again.
 
I'm not familiar with that acronym.

In the USA, we use this acronym: PEMDAS

The mnemonic is: "Please Excuse My Dear Aunt Sally".


P - parentheses (i.e., grouping symbols)

E - exponentiation

MD - multiplication/division

AS - addition/subtraction


I'm concerned that you wrote "and finally subtraction".

Addition and subtraction are done, in order as they appear, from left to right.

Multiplication and division are done, in order as they appear, from left to right.



Your solution has a sign error on x = 3/2.


\(\displaystyle \dfrac{-1 \pm \sqrt{1 + 24}}{2(2)}\)

\(\displaystyle \dfrac{-1 \pm 5}{4}\)

\(\displaystyle \dfrac{-1 + 5}{4} \text{OR} \dfrac{-1 - 5}{4}\)

\(\displaystyle \dfrac{4}{4} \text{OR} \dfrac{-6}{4}\)

I would have thought that x = -3/2 as your solution also says; -6/4 imples -1 1/2 = -3/2

Would you agree.

With regards the mnemonic BEDMAS, although there are different ways of writing it, it is actually the same.

B = Brackets
E = Exponents
D = Division
M = multiplication
A = Addition
S = Subtraction

So if I put them in order as above, B = 1 and S = 6, then working down the list 5/4 by division before adding or subtracting = wrong answers.

I understand what you pointed above about working from left to right in order of appearance, but can you see where my confusion occurs with how I have been informed to use BEDMAS?

Regards

Probability
 
I would have thought that x = -3/2 as your solution also says

But, you previously typed 3/2.



With regards the mnemonic BEDMAS, although there are different ways of writing it, it is actually the same.

Apparently so, as you previously typed BIDMAS. ;)

I was able to guess B, but I could not guess I.



So if I [number the letters in the acronym] as above, B = 1 and S = 6, then working down the list 5/4 by division before adding or subtracting = wrong answers.

you see where my confusion occurs with how I have been informed to use BEDMAS?

Yikes! I sure hope that it was not your teacher who informed you that way. That's terrible!
 
YES, and I'll repeat it!
Was not commenting on your "learning speed", but on posting clarity.


We're now up to 20 posts in this thread; if you had simply posted
properly, this would have been over with within 2 or 3 posts:
remember that there are other students asking questions, not just you!

AND you were told right at the start:
"Are you getting sloppy? You've posted 59 times already.
That's not an equation; all equations contain an equal sign.
Did you leave off your grouping symbols?"

Well, have a good day...

I made a mistake by not putting the equals sign there, my fault.

2/(x+1) + 2(x) - 3 / x = 0

That's not an [URL="http://freemathhelp.com/equations.html"]equation[/URL];

Yes it is :mrgreen::mrgreen:
 
But, you previously typed 3/2.





Apparently so, as you previously typed BIDMAS. ;)

I was able to guess B, but I could not guess I.





Yikes! I sure hope that it was not your teacher who informed you that way. That's terrible!

I = indices, exponents, powers.;)

Please advise what is the correct order of using BEDMAS

Regards

Prob
 
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