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Which of the following is equivalent to [1 - {1/(x + 1)]/x for all

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Which of the following is equivalent to [1 - {1/(x + 1)]/x for all  [#permalink]

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New post 15 Jan 2018, 06:12
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Which of the following is equivalent to \(\frac{1 - \frac{1}{x + 1}}{x}\) for all values of x for which both expressions are defined?

(A) 1
(B) x + 1
(C) 1/x
(D) 1/(x + 1)
(E) x^2 + x

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Which of the following is equivalent to [1 - {1/(x + 1)]/x for all  [#permalink]

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New post 15 Jan 2018, 07:52
Bunuel wrote:
Which of the following is equivalent to \(\frac{1 - \frac{1}{x + 1}}{x}\) for all values of x for which both expressions are defined?

(A) 1
(B) x + 1
(C) 1/x
(D) 1/(x + 1)
(E) x^2 + x


Method 1: Algebraic

\(\frac{1 - \frac{1}{x + 1}}{x}=\frac{x+1-1}{(x+1)x}\)

\(\frac{x}{(x+1)x}=\frac{1}{x+1}\)

Option D

-----------------------------

Method 2: Chose smart number for \(x\). let \(x=1\)

therefore \(\frac{1 - \frac{1}{x + 1}}{x}=1-\frac{1}{2}=\frac{1}{2}\)

Substitute \(x=1\) in options and only option D\(=\frac{1}{2}\)
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Re: Which of the following is equivalent to [1 - {1/(x + 1)]/x for all  [#permalink]

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New post 31 Mar 2018, 12:18
Bunuel wrote:
Which of the following is equivalent to \(\frac{1 - \frac{1}{x + 1}}{x}\) for all values of x for which both expressions are defined?

(A) 1
(B) x + 1
(C) 1/x
(D) 1/(x + 1)
(E) x^2 + x



I got this one by choosing smart number. But I don't know how to solve it algebraically. Can someone please write out the steps? Thanks.
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Which of the following is equivalent to [1 - {1/(x + 1)]/x for all  [#permalink]

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New post 31 Mar 2018, 20:34
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msurls wrote:
Bunuel wrote:
Which of the following is equivalent to \(\frac{1 - \frac{1}{x + 1}}{x}\) for all values of x for which both expressions are defined?

(A) 1
(B) x + 1
(C) 1/x
(D) 1/(x + 1)
(E) x^2 + x

I got this one by choosing smart number. But I don't know how to solve it algebraically. Can someone please write out the steps? Thanks.

msurls , sure. (Just making sure that you saw niks18 's solution.) Complex fractions can be a pain. \(\frac{1 - \frac{1}{x + 1}}{x}\)

One method (and there are many): simplify the numerator first, then the whole fraction
If there were a fraction in the denominator (instead of in the numerator, as here),
you could simplify the denominator first.
Fractions in both numerator and denominator?
Typically, use LCM of all fractions in the complex fraction, and
multiply that LCM "through" all terms in numerator and denominator.

Here:

1) Simplify numerator ONLY. Use LCM to convert terms into like fractions. Numerator:
\(1 -\frac{1}{x+1}\)
LCM of 1 and (x+1) is (x+1)

Second term has correct denominator. Multiply first term by \(\frac{LCM}{LCM}\)(equivalent of 1)

\((\frac{x+1}{x+1} * 1) - \frac{1}{x+1}\) =

\(\frac{x+1}{x+1} - \frac{1}{x+1}\)

2) Subtract (denominator does not change)
\((\frac{x+1}{x+1} - \frac{1}{x+1}) =\frac{(x+1)-1}{x+1}=\frac{x}{x+1}\)

3) Put that whole expression, the new numerator, over the original denominator, \(x\)

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

4) Express division as multiplication* and multiply
\(\frac{(\frac{x}{x+1})}{x} =(\frac{x}{x+1}*\frac{1}{x})= \frac{x*1}{(x+1)*x}=\frac{x}{x(x+1)}\)

5) Simplify. Factor out ("cancel") \(x\) in numerator and denominator:
\(\frac{x}{x(x+1)}=\frac{1}{(x+1)}\)

Answer D

* Division as multiplication

\(\frac{(\frac{a}{b})}{c}=\frac{a}{b}*\frac{1}{c}=\frac{a}{bc}\)

In short, \(\frac{(\frac{a}{b})}{c}=\frac{a}{b*c}\)


Hope that helps.
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