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OpenStudy (anonymous):

Add and simplify x+2x/x+1

OpenStudy (zzr0ck3r):

\(\frac{x+2x}{x+1}\)?

OpenStudy (anonymous):

A. 3x/x+1 B. x(x+3)/x+1 C. 4x+1/x+1 D. 2x^2/x+1 would it be D?

OpenStudy (anonymous):

@zzr0ck3r

OpenStudy (zzr0ck3r):

\(\frac{x+2x}{x+1}\) \(\huge ?\)

OpenStudy (anonymous):

A

OpenStudy (anonymous):

Does x+2x/x+1=(A.) 3x/x+1

OpenStudy (zzr0ck3r):

please answer my question.....x+2x/x+1 is very ambiguous. I dont know if you mean \(x+\frac{2x}{x}+1\) or \(\frac{x+2x}{x}+1\), or \(\frac{x+2x}{x+1}\). So which one is it?

OpenStudy (zzr0ck3r):

i.e use parenthesis

OpenStudy (anonymous):

the 1st one

OpenStudy (zzr0ck3r):

\(x+\frac{2x}{x}+1\)?

OpenStudy (anonymous):

yes

OpenStudy (zzr0ck3r):

\(x+\frac{2x}{x}+1=x+2+1=x+3\) this is not one of your options, so im guessing you did NOT mean the 1st one.....

OpenStudy (zzr0ck3r):

the only one that makes sense to your answers is \(\frac{x+2x}{x+1}\)

OpenStudy (zzr0ck3r):

how does that look like the first one?

OpenStudy (zzr0ck3r):

\(x+\frac{2x}{x+1}\) does not look like \(x+\frac{2x}{x}+1\) to me:P

OpenStudy (zzr0ck3r):

ok \(x+\frac{2x}{x+1}\) we need a common denominator which will be \(x+1\) so multiple the first term (top and bottom) by (x+1) \(x+\frac{2x}{x+1}=x\frac{x+1}{x+1}+\frac{2x}{x+1}=\frac{x(x+1)}{x+1}+\frac{2x}{x+1}=\frac{x^2+x}{x+1}+\frac{2x}{x+1}=\frac{x^2+x+2x}{x+1}=\\\frac{x^2+3x}{x+1}=x\frac{x+3}{x+1}\)

OpenStudy (anonymous):

Thank you

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