Equation solving/solving for x

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Homework Help Overview

The problem involves solving the equation 1 / (2x + 1) − 2x / (x − 3) = (x + 1) / (2x^2 − 5x − 3) for the variable x. Participants are attempting to manipulate the equation and express it in a more manageable form.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are discussing the need for additional parentheses in the equation to clarify its structure. Some express confusion about the current form and its implications, while others suggest rewriting the equation to facilitate solving it. There are mentions of using software tools like Maple to check solutions.

Discussion Status

Several participants have provided feedback on the original equation's formatting, indicating that clarity is essential for further progress. Suggestions have been made to consolidate terms and consider the implications of the denominator being zero. There is an ongoing exploration of how to manipulate the equation effectively.

Contextual Notes

Participants note that the denominator becomes zero for specific values of x, which raises questions about the validity of those solutions. There is also an emphasis on ensuring that all steps are shown clearly to avoid misunderstandings.

Weightofananvil
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Homework Statement


1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3

solve for x. check solution in original equation

The Attempt at a Solution



I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.

I really don't know where to go or what to do.
I even put the answer into maple and tried to work it backwards.
Super stuck. thanks for the help in advance.
 
Last edited by a moderator:
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Maybe it is not written correctly.
 
Weightofananvil said:

Homework Statement


1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3

solve for x. check solution in original equation

The Attempt at a Solution



I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.

I really don't know where to go or what to do.
I even put the answer into maple and tried to work it backwards.
Super stuck. thanks for the help in advance.

I guess you omitted a few couples of parentheses .
What you wrote is equivalent to 0.5 x -2 -3 =x+0.5 x^2-5x-3 - are you sure you wanted to write that?
 
Last edited by a moderator:
Weightofananvil said:

Homework Statement


1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3
As ehild already said, you need quite a few more parentheses here.

What you wrote is properly interpreted to mean this:
##\frac 1 2 x + 1 - \frac{2x}{x} - 3 = x + \frac 1 2 x^2 - 5x - 3##
Weightofananvil said:
solve for x. check solution in original equation

The Attempt at a Solution



I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.
Again, you need some parentheses, plus show your intervening steps in getting to the above.

Regarding parentheses, when you write fractions on a single line, if the numerator or denominator has more than a single term, put parentheses around it.
For example: ##\frac{x^2 + 3x + 2}{x + 1}## should be written as (x^2 + 3x + 2)/(x + 1), if written on a single line.
Weightofananvil said:
I really don't know where to go or what to do.
I even put the answer into maple and tried to work it backwards.
Super stuck. thanks for the help in advance.
 
Weightofananvil said:

Homework Statement


1 / 2x + 1 − 2x / x − 3 = x + 1 / 2x ^ 2 − 5x − 3

solve for x. check solution in original equation

The Attempt at a Solution



I've tried working this out a couple different ways. All I can change it to is 4x^2 - x - 3 / 2x^2 - 5x - 3 = x+1/2x^2-5x-3.

I really don't know where to go or what to do.
I even put the answer into maple and tried to work it backwards.
Super stuck. thanks for the help in advance.
Using parentheses, (4x^2- x- 3)/(2x^2- 5x- 3)= (x+ 1)/(2x^2- 5x- 3)
Note that the denominator is 0 for x= 3 and x= -1/2. As long as the denominator is NOT 0, we can multiply on both sides by it to get
4x^2- x- 3= x+ 1 which is the same as the quadratic equation 4x^2- 2x- 4= 0
 
From where you are you could put all on the same side since you have a common denominator you can put it away and remains a quadratic equation
 
Thanks for the help everyone!
 

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