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If x is different from 3 and x is different from 6, then
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20 Apr 2019, 10:11
20 Apr 2019, 10:11
Expert Reply
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Question Stats:
88% (01:46) correct
11% (02:34) wrong based on 18 sessions
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If \(x \neq 3\) and \(x \neq 6\), then \(\frac{2x^2-72}{x-6} - \frac{2x^2-18}{x-3}=\)
(A) 3
(B) 6
(C) 9
(D) 12
(E) 15
(A) 3
(B) 6
(C) 9
(D) 12
(E) 15
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Re: If x is different from 3 and x is different from 6, then
[#permalink]
21 Apr 2019, 06:36
21 Apr 2019, 06:36
1
Carcass wrote:
If \(x \neq 3\) and \(x \neq 6\), then \(\frac{2x^2-72}{x-6} - \frac{2x^2-18}{x-3}=\)
(A) 3
(B) 6
(C) 9
(D) 12
(E) 15
(A) 3
(B) 6
(C) 9
(D) 12
(E) 15
\(\frac{2x^2-72}{x-6} - \frac{2x^2-18}{x-3}=\frac{2(x^2-36)}{x-6} - \frac{2(x^2-9)}{x-3}\)
\(=\frac{2(x+6)(x-6)}{x-6} - \frac{2(x+3)(x-3)}{x-3}\)
\(=2(x+6)- 2(x+3)\)
\(=2x+12-2x-6\)
\(=6\)
Answer: B
Cheers,
Brent
gmatclubot
Re: If x is different from 3 and x is different from 6, then [#permalink]
21 Apr 2019, 06:36
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