If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
The OA is the option B.
Could someone tell me what is the best way of solving this PS question? Should I try number by number? I'd be thankful for your help.
If |x - 2| = |x + 3|, x could equal
This topic has expert replies
-
- Legendary Member
- Posts: 2898
- Joined: Thu Sep 07, 2017 2:49 pm
- Thanked: 6 times
- Followed by:5 members
Hello M7MBA.
Well, you can plug in the values and see which option is the correct.
Also, we can solve it by expanding the absolute values |x - 2| and |x + 3| as follows:
Each expression between the absolute values is equal to zero when x=2 and x=-3. Then, we have the following cases:
Case 1: x<-3.
- If x<-3 then x+3<0.
- If x<-3 then x<2 and therefore x-2<0.
Hence $$\left|x-2\right|=\left|x+3\right|\ \ \Leftrightarrow\ \ \ -\left(x-2\right)=-\left(x+3\right)\ \ \ \Leftrightarrow\ \ -x+2=-x-3\ \ \ \Leftrightarrow\ \ 2=-3\ \ \ False.$$ In this case we didn't get a real solution.
Case 2: -3 < x < 2.
- If -3<x then x+3>0.
- If x < 2then x-2<0.
Hence $$\left|x-2\right|=\left|x+3\right|\ \ \Leftrightarrow\ \ \ -\left(x-2\right)=x+3\ \ \ \Leftrightarrow\ \ -x+2=x+3\ \ \ \Leftrightarrow\ \ 2x=-1\ \ \ \Leftrightarrow\ \ \ x=-\frac{1}{2}.$$ This implies that the correct answer is the option [spoiler]C=-1/2[/spoiler].
Finally
Case 3: x > 2
- If x > 2 then x-2 > 0.
- If x > 2 then x > -3 and therefore x+3 > 0.
Hence $$\left|x-2\right|=\left|x+3\right|\ \ \Leftrightarrow\ \ \ x-2=x+3\ \ \ \Leftrightarrow\ \ -2=3\ \ False$$ I hope it helps you.
Well, you can plug in the values and see which option is the correct.
Also, we can solve it by expanding the absolute values |x - 2| and |x + 3| as follows:
Each expression between the absolute values is equal to zero when x=2 and x=-3. Then, we have the following cases:
Case 1: x<-3.
- If x<-3 then x+3<0.
- If x<-3 then x<2 and therefore x-2<0.
Hence $$\left|x-2\right|=\left|x+3\right|\ \ \Leftrightarrow\ \ \ -\left(x-2\right)=-\left(x+3\right)\ \ \ \Leftrightarrow\ \ -x+2=-x-3\ \ \ \Leftrightarrow\ \ 2=-3\ \ \ False.$$ In this case we didn't get a real solution.
Case 2: -3 < x < 2.
- If -3<x then x+3>0.
- If x < 2then x-2<0.
Hence $$\left|x-2\right|=\left|x+3\right|\ \ \Leftrightarrow\ \ \ -\left(x-2\right)=x+3\ \ \ \Leftrightarrow\ \ -x+2=x+3\ \ \ \Leftrightarrow\ \ 2x=-1\ \ \ \Leftrightarrow\ \ \ x=-\frac{1}{2}.$$ This implies that the correct answer is the option [spoiler]C=-1/2[/spoiler].
Finally
Case 3: x > 2
- If x > 2 then x-2 > 0.
- If x > 2 then x > -3 and therefore x+3 > 0.
Hence $$\left|x-2\right|=\left|x+3\right|\ \ \Leftrightarrow\ \ \ x-2=x+3\ \ \ \Leftrightarrow\ \ -2=3\ \ False$$ I hope it helps you.
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Case 1: signs unchangedM7MBA wrote:If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
x-2 = x+3
-2 = 3
Since the resulting equation is invalid, Case 1 is not possible.
Case 2: signs changed in ONE of the absolute values
x-2 = -x-3
2x = -1
x = -1/2.
The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Alternate approach:M7MBA wrote:If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
|a-b| = the distance between a and b.
|a+b| = |a-(-b)| = the distance between a and -b.
|x - 2| = |x + 3| implies the following:
The distance between x and 2 is equal to the distance between x and -3.
In other words, x must be HALFWAY BETWEEN -3 AND 2.
Halfway between two numbers = the AVERAGE of the two numbers:
(-3+2)/2 = -1/2.
the correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
There are 3 steps to solving equations involving ABSOLUTE VALUE:M7MBA wrote:If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug solutions into original equation to check for extraneous roots
Given: |x - 2| = |x + 3|
case a: x - 2 = x + 3
Subtract x from both sides to get: -2 = 3
No good.
So, this equation has no solution.
case b: x - 2 = -(x + 3)
Simplify: x - 2 = -x - 3
Add x to both sides: 2x - 2 = -3
Add 2 to both sides: 2x = -1
Solve: x = -1/2
IMPORTANT: Plug x = -1/2 into the ORIGINAL equation, to make sure it isn't an extraneous root.
We get: |-1/2 - 2| = |-1/2 + 3|
Evaluate: |-2.5| = |2.5|
Evaluate: 2.5 = 2.5
Works!
So, x = -1/2 IS a solution.
Answer: B
Cheers,
Brent
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
The warning above must be heeded when only one side of an equation is an absolute value.IMPORTANT: Plug x = -1/2 into the ORIGINAL equation, to make sure it isn't an extraneous root.
But when BOTH sides of an equation are absolute values, any solution that we derive will be guaranteed to work, so the step above is unnecessary.
Since both sides of |x - 2| = |x + 3| are absolute values, we do not need to check the validity of x=-1/2.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
- Jeff@TargetTestPrep
- GMAT Instructor
- Posts: 1462
- Joined: Thu Apr 09, 2015 9:34 am
- Location: New York, NY
- Thanked: 39 times
- Followed by:22 members
By scanning the answer choices, we see that when x = -1/2, we haveM7MBA wrote:If |x - 2| = |x + 3|, x could equal
A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution
|-1/2 - 2| = |-5/2| = 5/2 and |-1/2 + 3| = |5/2| = 5/2
Alternate Solution:
We consider two cases for this absolute value question.
Case 1. Drop the absolute value symbols and solve for x.
x - 2 = x + 3
0 = 5
This doesn't work.
Case 2. (x - 2) is positive and (x + 3) is negative.
x - 2 = -(x + 3)
x - 2 = -x - 3
2x = -1
x = -½
When x = -1/2, we see that |x - 2| = |x + 3| becomes |(-1/2) - 2| = |(-1/2) + 3|, or 5/2 = 5/2, which is a true statement. Thus, the solution for this equation is x = -1/2.
Answer: B
Jeffrey Miller
Head of GMAT Instruction
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews