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jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,296
Given Kudos: 355
Concentration: Finance
Schools: Wharton '17 (A) HBS '17 (WA$)
A
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Difficulty:
15%
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Question Stats:
82% (01:44) correct
18%
(02:12)
wrong
based on 395
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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
(A) 2
(B) 1
(C) -1
(D) -2
(E) -3
11 Oct 2013, 17:29
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jlgdr
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.
jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,296
Own Kudos:
Given Kudos: 355
Concentration: Finance
Schools: Wharton '17 (A) HBS '17 (WA$)
11 Oct 2013, 17:49
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Bunuel
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?
