GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 10 Apr 2020, 02:44

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

(x – 2)(x + 4) – (x – 3)(x – 1) = 0

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 62696
(x – 2)(x + 4) – (x – 3)(x – 1) = 0  [#permalink]

Show Tags

New post Updated on: 10 Jul 2019, 22:48
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

87% (01:37) correct 13% (02:07) wrong based on 60 sessions

HideShow timer Statistics


Originally posted by Bunuel on 17 Apr 2019, 20:52.
Last edited by SajjadAhmad on 10 Jul 2019, 22:48, edited 1 time in total.
Added Source
Intern
Intern
avatar
B
Joined: 06 Apr 2019
Posts: 7
Re: (x – 2)(x + 4) – (x – 3)(x – 1) = 0  [#permalink]

Show Tags

New post 18 Apr 2019, 03:32
The approach is simple, just use the FOIL method and then subtract both the sides.

FOIL for (x-2)(x+4) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8
FOIL for (x-3)(x-1) = x^2 - x - 3x + 3 = x^2 - 4x + 3

Substituting them in the equation

(x^2 + 2x - 8) - (x^2 - 4x + 3) = 0
6x - 11 = 0
6x = 11
x = 11 / 6

Hence the answer is option (C)
Intern
Intern
avatar
B
Joined: 11 Apr 2019
Posts: 14
Schools: IE MiF "21 (A)
Re: (x – 2)(x + 4) – (x – 3)(x – 1) = 0  [#permalink]

Show Tags

New post 19 Apr 2019, 04:34
danz1ka19 wrote:
The approach is simple, just use the FOIL method and then subtract both the sides.

FOIL for (x-2)(x+4) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8
FOIL for (x-3)(x-1) = x^2 - x - 3x + 3 = x^2 - 4x + 3

Substituting them in the equation

(x^2 + 2x - 8) - (x^2 - 4x + 3) = 0
6x - 11 = 0
6x = 11
x = 11 / 6

Hence the answer is option (C)


You mean E, right? :)
Intern
Intern
avatar
B
Joined: 06 Apr 2019
Posts: 7
Re: (x – 2)(x + 4) – (x – 3)(x – 1) = 0  [#permalink]

Show Tags

New post 19 Apr 2019, 11:26
vargamartin1031 wrote:
danz1ka19 wrote:
The approach is simple, just use the FOIL method and then subtract both the sides.

FOIL for (x-2)(x+4) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8
FOIL for (x-3)(x-1) = x^2 - x - 3x + 3 = x^2 - 4x + 3

Substituting them in the equation

(x^2 + 2x - 8) - (x^2 - 4x + 3) = 0
6x - 11 = 0
6x = 11
x = 11 / 6

Hence the answer is option (C)


You mean E, right? :)


Oh yeah! my bad lol.
Director
Director
User avatar
V
Joined: 27 May 2012
Posts: 972
Re: (x – 2)(x + 4) – (x – 3)(x – 1) = 0  [#permalink]

Show Tags

New post 23 Apr 2019, 11:49
Bunuel wrote:
(x – 2)(x + 4) – (x – 3)(x – 1) = 0

(A) –5
(B) –1
(C) 0
(D) 1/2
(E) 11/6


Are we supposed to find the value of x, in this question ?
_________________
- Stne
Manager
Manager
avatar
P
Joined: 26 Mar 2019
Posts: 115
Premium Member Reviews Badge
Re: (x – 2)(x + 4) – (x – 3)(x – 1) = 0  [#permalink]

Show Tags

New post 02 May 2019, 04:48
stne wrote:
Bunuel wrote:
(x – 2)(x + 4) – (x – 3)(x – 1) = 0

(A) –5
(B) –1
(C) 0
(D) 1/2
(E) 11/6


Are we supposed to find the value of x, in this question ?


Yes. Just simplifying the equation
Intern
Intern
avatar
Joined: 02 May 2019
Posts: 6
Re: (x – 2)(x + 4) – (x – 3)(x – 1) = 0  [#permalink]

Show Tags

New post 02 May 2019, 04:56
(x – 2)(x + 4) – (x – 3)(x – 1) = 0
the answer is 5 to this question
GMAT Club Bot
Re: (x – 2)(x + 4) – (x – 3)(x – 1) = 0   [#permalink] 02 May 2019, 04:56
Display posts from previous: Sort by

(x – 2)(x + 4) – (x – 3)(x – 1) = 0

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne