Simplifying Complex Expressions

4hannah

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Joined
Apr 13, 2021
Messages
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I’m hoping this question is in the right spot?

(if not, I apologize)

Write x/2 - 2x+4/x+1 as a single fraction.

my approach/steps are as follows:
After finding lowest common denominator,
(2)(x+1),

x(x+1) - 2(2x+4)/(2)(x+1)
= x^2 + x - 4x -8/ 2(x+1)
= x^2 -3x -8/2(x+1)
I know I’m missing something; is this the simplest fractional form?
 
Write x/2 - 2x+4/x+1 as a single fraction.

my approach/steps are as follows:
After finding lowest common denominator,
(2)(x+1),

x(x+1) - 2(2x+4)/(2)(x+1)
= x^2 + x - 4x -8/ 2(x+1)
= x^2 -3x -8/2(x+1)
I know I’m missing something; is this the simplest fractional form?
It appears that you omitted some parentheses that are needed to make this mean what you meant. I'll add them for you:

Write x/2 - (2x+4)/(x+1) as a single fraction.​
my approach/steps are as follows:
After finding lowest common denominator,
(2)(x+1),
[x(x+1) - 2(2x+4)]/[(2)(x+1)]
= [x^2 + x - 4x -8]/[2(x+1)]
= [x^2 -3x -8]/[2(x+1)]

Assuming that is what you meant, you are right so far.

To see if you can go further, you check whether the quadratic numerator can be factored; it can't, so you are finished.

Good work.
 
It appears that you omitted some parentheses that are needed to make this mean what you meant. I'll add them for you:

Write x/2 - (2x+4)/(x+1) as a single fraction.​
my approach/steps are as follows:
After finding lowest common denominator,
(2)(x+1),
[x(x+1) - 2(2x+4)]/[(2)(x+1)]
= [x^2 + x - 4x -8]/[2(x+1)]
= [x^2 -3x -8]/[2(x+1)]

Assuming that is what you meant, you are right so far.

To see if you can go further, you check whether the quadratic numerator can be factored; it can't, so you are finished.

Good work.
Thank you so much for the reply and your help (yep, parentheses are definitely in order!)

Thank you for confirming my work.

Kind regards
 
I’m hoping this question is in the right spot?

(if not, I apologize)

Write x/2 - 2x+4/x+1 as a single fraction.

my approach/steps are as follows:
After finding lowest common denominator,
(2)(x+1),

x(x+1) - 2(2x+4)/(2)(x+1)
= x^2 + x - 4x -8/ 2(x+1)
= x^2 -3x -8/2(x+1)
I know I’m missing something; is this the simplest fractional form?
What if the problem really was [math]\dfrac{x}{2} - 2x +\dfrac{4}{x} + 1[/math], don't you see that you would have written x/2 - 2x+4/x+1? Doesn't that trouble you? After all, writing two very different expressions the same way just can't be correct.
 
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