If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= : GMAT Problem Solving (PS)
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# If (z+3)/(z-1) + (z+1)/(z-3)=2, then z=

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Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
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Kudos [?]: 592 [0], given: 355

If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= [#permalink]

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11 Oct 2013, 16:10
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79% (02:08) correct 21% (01:11) wrong based on 107 sessions

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If $$\frac{z+3}{z-1}+\frac{z+1}{z-3} = 2$$, then z=

(A) 2
(B) 1
(C) -1
(D) -2
(E) -3
[Reveal] Spoiler: OA

Last edited by jlgdr on 13 Oct 2013, 04:01, edited 3 times in total.
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Joined: 02 Sep 2009
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Re: If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= [#permalink]

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11 Oct 2013, 16:29
jlgdr wrote:
If $$\frac{z+3}{z-1}+\frac{z+1}{z-3} = 2$$, then z=

(A) 2
(B) 1
(C) -1
(D) -2
(E) -3

I would solve this question by substituting the options but below is an algebraic approach if interested:

$$\frac{z+3}{z-1}+\frac{z+1}{z-3} = 2$$;

$$\frac{z^2-9+z^2-1}{(z-1)(z-3)}= 2$$;

$$2z^2-10=2z^2-8z+6$$;

$$8z=16$$;

$$z=2$$.

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Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 62

Kudos [?]: 592 [0], given: 355

Re: If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= [#permalink]

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11 Oct 2013, 16:49
Bunuel wrote:
jlgdr wrote:
If $$\frac{z+3}{z-1}+\frac{z+1}{z-3} = 2$$, then z=

(A) 2
(B) 1
(C) -1
(D) -2
(E) -3

I would solve this question by substituting the options but below is an algebraic approach if interested:

$$\frac{z+3}{z-1}+\frac{z+1}{z-3} = 2$$;

$$\frac{z^2-9+z^2-1}{(z-1)(z-3)}= 2$$;

$$2z^2-10=2z^2-8z+6$$;

$$8z=16$$;

$$z=2$$.

I agree, best approach is to sustitute. Now next question would be? Do we have a way to do a fast check and see which value to substitute first?
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Kudos [?]: 8 [1] , given: 45

Re: If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= [#permalink]

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12 Oct 2013, 21:16
1
KUDOS
jlgdr wrote:
Bunuel wrote:
jlgdr wrote:
If $$\frac{z+3}{z-1}+\frac{z+1}{z-3} = 2$$, then z=

(A) 2
(B) 1
(C) -1
(D) -2
(E) -3

I would solve this question by substituting the options but below is an algebraic approach if interested:

$$\frac{z+3}{z-1}+\frac{z+1}{z-3} = 2$$;

$$\frac{z^2-9+z^2-1}{(z-1)(z-3)}= 2$$;

$$2z^2-10=2z^2-8z+6$$;

$$8z=16$$;

$$z=2$$.

I agree, best approach is to sustitute. Now next question would be? Do we have a way to do a fast check and see which value to substitute first?

I agree...OA is A someone kindly change the OA in the question...
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
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Kudos [?]: 592 [0], given: 355

Re: If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= [#permalink]

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13 Oct 2013, 04:02
OA has been changed

Cheers

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Re: If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= [#permalink]

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21 Jul 2015, 08:43
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Re: If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= [#permalink]

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22 Jul 2015, 07:25
If z+3z−1+z+1z−3=2, then z=

(A) 2
(B) 1
(C) -1
(D) -2
(E) -3
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If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= [#permalink]

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27 Jul 2015, 00:23
OptimusPrepJanielle wrote:
If z+3z−1+z+1z−3=2, then z=

(A) 2
(B) 1
(C) -1
(D) -2
(E) -3

When you multiply with (z-1) and (z-3) in the first step, how does the denom. become (z-1)*(z-3)?
If (z+3)/(z-1) + (z+1)/(z-3)=2, then z=   [#permalink] 27 Jul 2015, 00:23
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