Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 Apr 2016, 04:10

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# What is wrong with this solution?

Author Message
TAGS:

### Hide Tags

Intern
Joined: 09 Oct 2011
Posts: 10
Location: United States
GMAT Date: 12-27-2011
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 5 [0], given: 6

What is wrong with this solution? [#permalink]

### Show Tags

04 Mar 2012, 21:49
I used two different methods to solve an equation and I'm getting different answers. I'm not sure where am I going wrong.

Method 1 - Solving for x
$$x^2-16=x-4$$
$$x^2-x-12=0$$
$$(x-4)(x+3)=0$$
$$x=4, -3$$

Method 2 - Solving for x
$$x^2-16=x-4$$
$$x^2-4^2=x-4$$
$$(x+4)(x-4)=x-4$$
$$(x+4)=\frac{(x-4)}{(x-4)}$$
$$(x+4)=1$$
$$x=-3$$

It looks like something is wrong with method 2. Also in method 2 x cannot be 4 else the denominator will be 0.
Can someone tell me what am I doing wrong in method 2.

Thank you
Math Expert
Joined: 02 Sep 2009
Posts: 32530
Followers: 5620

Kudos [?]: 68186 [1] , given: 9797

Re: What is wrong with this solution? [#permalink]

### Show Tags

05 Mar 2012, 01:25
1
KUDOS
Expert's post
vinayanand wrote:
I used two different methods to solve an equation and I'm getting different answers. I'm not sure where am I going wrong.

Method 1 - Solving for x
$$x^2-16=x-4$$
$$x^2-x-12=0$$
$$(x-4)(x+3)=0$$
$$x=4, -3$$

Method 2 - Solving for x
$$x^2-16=x-4$$
$$x^2-4^2=x-4$$
$$(x+4)(x-4)=x-4$$
$$(x+4)=\frac{(x-4)}{(x-4)}$$
$$(x+4)=1$$
$$x=-3$$

It looks like something is wrong with method 2. Also in method 2 x cannot be 4 else the denominator will be 0.
Can someone tell me what am I doing wrong in method 2.

Thank you

Second approach is wrong because what you are actually doing there is reducing (dividing) both parts of the equation by x-4, thus assuming with no ground for this that x-4 does not equal to zero (otherwise dividing by it wouldn't be allowed) hence you are excluding perfectly valid solution x-4=0 --> x=4.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

Hope it's clear.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6473
Location: Pune, India
Followers: 1750

Kudos [?]: 10449 [1] , given: 205

Re: What is wrong with this solution? [#permalink]

### Show Tags

05 Mar 2012, 05:02
1
KUDOS
Expert's post
vinayanand wrote:
I used two different methods to solve an equation and I'm getting different answers. I'm not sure where am I going wrong.

Method 1 - Solving for x
$$x^2-16=x-4$$
$$x^2-x-12=0$$
$$(x-4)(x+3)=0$$
$$x=4, -3$$

Method 2 - Solving for x
$$x^2-16=x-4$$
$$x^2-4^2=x-4$$
$$(x+4)(x-4)=x-4$$
$$(x+4)=\frac{(x-4)}{(x-4)}$$
$$(x+4)=1$$
$$x=-3$$

It looks like something is wrong with method 2. Also in method 2 x cannot be 4 else the denominator will be 0.
Can someone tell me what am I doing wrong in method 2.

Thank you

Check out this post for an explanation on when you can/cannot divide by a variable:
in-which-one-of-the-following-choices-must-p-be-greater-than-127621.html#p1048624
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Intern
Joined: 09 Oct 2011
Posts: 10
Location: United States
GMAT Date: 12-27-2011
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 5 [0], given: 6

Re: What is wrong with this solution? [#permalink]

### Show Tags

05 Mar 2012, 05:58
Thank you Bunuel and Karishma
Re: What is wrong with this solution?   [#permalink] 05 Mar 2012, 05:58
Similar topics Replies Last post
Similar
Topics:
I think the "correct" anwser in the solutions is wrong for this one. W 1 12 Dec 2015, 07:06
3 What is the range of solutions for |x^2 - 4| > 3x ? 7 05 Nov 2014, 11:29
Not for solution but for suggestion 2 08 Mar 2011, 12:19
1 High School Math - What am i doing wrong here? 4 14 Aug 2010, 04:43
105 Pumps A, B, and C operate at their respective constant rates. Pumps A 19 10 Nov 2009, 15:14
Display posts from previous: Sort by