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# If x ≠ 3 and x ≠ 6, then (2x^2 - 72)/(x - 6) – (2x^2 -18)/(x - 3) =

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Math Expert
Joined: 02 Sep 2009
Posts: 58454
If x ≠ 3 and x ≠ 6, then (2x^2 - 72)/(x - 6) – (2x^2 -18)/(x - 3) =  [#permalink]

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10 Apr 2019, 21:51
00:00

Difficulty:

15% (low)

Question Stats:

94% (01:18) correct 6% (01:29) wrong based on 17 sessions

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If x ≠ 3 and x ≠ 6, then $$\frac{2x^2 - 72}{x - 6} – \frac{2x^2 -18}{x - 3} =$$

(A) 3
(B) 6
(C) 9
(D) 12
(E) 15

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Joined: 18 Aug 2017
Posts: 5020
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: If x ≠ 3 and x ≠ 6, then (2x^2 - 72)/(x - 6) – (2x^2 -18)/(x - 3) =  [#permalink]

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11 Apr 2019, 00:51
Bunuel wrote:
If x ≠ 3 and x ≠ 6, then $$\frac{2x^2 - 72}{x - 6} – \frac{2x^2 -18}{x - 3} =$$

(A) 3
(B) 6
(C) 9
(D) 12
(E) 15

simplify expression
$$\frac{2x^2 - 72}{x - 6} – \frac{2x^2 -18}{x - 3} =$$

we get 2(x+6)-2( x+3)=0
x = 6

IMO B
Re: If x ≠ 3 and x ≠ 6, then (2x^2 - 72)/(x - 6) – (2x^2 -18)/(x - 3) =   [#permalink] 11 Apr 2019, 00:51
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