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If (z+3)/(z-1) + (z+1)/(z-3)=2, then z=

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

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New post 11 Oct 2013, 16:10
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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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Re: If (z+3)/(z-1) + (z+1)/(z-3)=2, then z= [#permalink]

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New post 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:

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

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

\(2z^2-10=2z^2-8z+6\);

\(8z=16\);

\(z=2\).

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

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New post 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:

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

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

\(2z^2-10=2z^2-8z+6\);

\(8z=16\);

\(z=2\).

Answer: A.


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

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New post 12 Oct 2013, 21:16
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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:

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

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

\(2z^2-10=2z^2-8z+6\);

\(8z=16\);

\(z=2\).

Answer: A.


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

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New post 13 Oct 2013, 04:02
OA has been changed

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

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

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New post 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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New post 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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