Equation solving/solving for x

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The discussion revolves around solving the equation 1/(2x + 1) - 2x/(x - 3) = (x + 1)/(2x^2 - 5x - 3). Participants emphasize the need for proper parentheses to clarify the equation's structure, as misinterpretation can lead to confusion. One user suggests rewriting the equation to isolate terms, leading to the quadratic equation 4x^2 - 2x - 4 = 0. They also note that the denominator cannot equal zero for valid solutions. Overall, the conversation highlights the importance of clarity in mathematical expressions when solving for x.
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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.
 
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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?
 
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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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