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# If |x - 2| = |x + 3|, x could equal

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Math Expert
Joined: 02 Sep 2009
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If |x - 2| = |x + 3|, x could equal  [#permalink]

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07 Mar 2016, 09:22
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If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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07 Mar 2016, 20:50
2

I back solved the question by putting answer choices in the question. -1/2 gives 5/2 on both sides.

Thanks
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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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07 Mar 2016, 20:58
I started drawing a number line, but then I thought better of it and just plugged in the values form the answers.
A) -5: LHS: |-5-2| = 7, RHS: |-5+3| = 2, no good
B) -1/2: LHS: |-1/2 - 2| = 2.5, RHS: |-1/2 + 3| = 2.5, Correct

Thus B
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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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07 Mar 2016, 21:04
2
2
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

By definition, |x-a| = distance of x from a. In other words, |x-2| = distance of x from 2 and |x+3| = distance of x from -3.

Thus, if |x-2| = |x+3| ---> distance of x from 2 = distance of x from -3. Thus, x = halfway between -3 and 2.

Distance between -3 and 2 = 5 units. Thus x is 2.5 units away from 2 or -3 , giving you x = 2-2.5 or = -3+2.5 = -0.5,

B is thus the correct answer.
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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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08 Mar 2016, 00:39
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

I just backsolved and was able to see that B was the answer for me. Both sides end up with the same result of 5/2.
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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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08 Mar 2016, 01:15
solving equation we get -1/2

Ans ; B
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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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11 Jun 2016, 11:18
1
1
|x - 2| = |x + 3|

We see both the sides are positive, hence we can square both the sides and we get

X^2 + 4 -4x = X^2 + 9 -6x

10x= -5

X= -1/2

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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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12 Jun 2016, 02:19
1
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

Since there are two Equations. Lets solve by Critical Value

|x - 2| = |x + 3| Critical Value will be 2, -3 hence we Have 3 Solutions

Solution No 1 When X < -3 ( RHS Will be Negative )

There fore x-2 = - (x + 3 ) => 2x = -1 => x = -1/2 which is not in line with x< -3 , hence rejected

Solution No 2 When -3 < X <= 2 ( LHS will be Negative )

-(x-2) = x+3 => 2x= -1 => x = -1/2 which fits in the range Hence Solution.

We can check by substituting in |x - 2| = |x + 3| Therefore 5/2= 5/2
Solution 3 when X > 2 both sides positive

x-2 = x+3 No Value hence rejected
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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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12 Jun 2016, 03:59
1
Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side
(x-2)^2= (x+3)^2
by solving we get x= -1/2 .
Please correct me if i missed something.
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Posts: 7563
Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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12 Jun 2016, 04:33
1
Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side
(x-2)^2= (x+3)^2
by solving we get x= -1/2 .
Please correct me if i missed something.
..

Hi sudhir,

yes, whenever you have ONLY mods on both sides..
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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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24 Jun 2016, 08:12
1
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

Given |x - 2| = |x + 3|

The best way to get rid of the modulus is to square on both sides.

$$(x - 2)^2$$ = $$(x+3)^2$$

=> $$x^2$$ + 4 + 4x = $$x^2$$ + 9 + 6x

=> x = -1/2.

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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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24 Jul 2016, 13:39
Razween wrote:
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

Since there are two Equations. Lets solve by Critical Value

|x - 2| = |x + 3| Critical Value will be 2, -3 hence we Have 3 Solutions

Solution No 1 When X < -3 ( RHS Will be Negative )

There fore x-2 = - (x + 3 ) => 2x = -1 => x = -1/2 which is not in line with x< -3 , hence rejected

Solution No 2 When -3 < X <= 2 ( LHS will be Negative )

-(x-2) = x+3 => 2x= -1 => x = -1/2 which fits in the range Hence Solution.

We can check by substituting in |x - 2| = |x + 3| Therefore 5/2= 5/2
Solution 3 when X > 2 both sides positive

x-2 = x+3 No Value hence rejected

In the first condition in which X is less than -3, why isn't the (x-2) term negative as well? The expression (x-2) is negative for all values where X is less than 2.
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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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25 Jul 2016, 01:18
astroshagger wrote:
Razween wrote:
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

Since there are two Equations. Lets solve by Critical Value

|x - 2| = |x + 3| Critical Value will be 2, -3 hence we Have 3 Solutions

Solution No 1 When X < -3 ( RHS Will be Negative )

There fore x-2 = - (x + 3 ) => 2x = -1 => x = -1/2 which is not in line with x< -3 , hence rejected

Solution No 2 When -3 < X <= 2 ( LHS will be Negative )

-(x-2) = x+3 => 2x= -1 => x = -1/2 which fits in the range Hence Solution.

We can check by substituting in |x - 2| = |x + 3| Therefore 5/2= 5/2
Solution 3 when X > 2 both sides positive

x-2 = x+3 No Value hence rejected

In the first condition in which X is less than -3, why isn't the (x-2) term negative as well? The expression (x-2) is negative for all values where X is less than 2.

Yes you are right.

If $$x < -3$$, then $$|x - 2| = -(x - 2)$$ and $$|x + 3| = -(x + 3)$$;

If $$-3 \leq x \leq 2$$, then $$|x - 2| = -(x - 2)$$ and $$|x + 3| = x + 3$$;

If $$x > 2$$, then $$|x - 2| = x - 2$$ and $$|x + 3| = x + 3$$.
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If |x - 2| = |x + 3|, x could equal  [#permalink]

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27 Jul 2017, 02:59
|x - 2| - |x + 3|=0
We substitute 2,-3 in the above expression accordingly and plot the same in number line.
Now minimum distance between the two points is 5 which is less than RHS. hence no real solution.
Option E.

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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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29 Jul 2017, 09:02
chetan2u wrote:
Bunuel, in one of your post , you ave explained if we have mod on both side of the equation, we can safely square on both side
(x-2)^2= (x+3)^2
by solving we get x= -1/2 .
Please correct me if i missed something.
..

Hi sudhir,

yes, whenever you have ONLY mods on both sides..

hi

so far I can remember, square rooting hides the positive/ negative issues. So, as the sign of x is unknown, it is safe to take cube roots...

please correct me if I am wrong ...

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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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29 Jul 2017, 09:14
ssislam wrote:

hi

so far I can remember, square rooting hides the positive/ negative issues. So, as the sign of x is unknown, it is safe to take cube roots...

please correct me if I am wrong ...

Hello ssislam, In this case, it will not hide as we are taking the square root (or rather assuming) the square root of x in the entire equation.

For some mod questions, you can take do a square on both sides and try to solve the question. Hope this helps
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Re: If |x - 2| = |x + 3|, x could equal  [#permalink]

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29 Jul 2017, 20:51
B is the right option
x-2=-x-3
2x=-3+2
2x=-1
x=-1/2
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If |x - 2| = |x + 3|, x could equal  [#permalink]

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30 Jul 2017, 10:16
1
Bunuel wrote:
If |x - 2| = |x + 3|, x could equal

A. -5
B. -1/2
C. 1/2
D. 5
E. No real solution

With absolute value equations this simple, for me, the method below is easy and fast (less than a minute).

The only way the two sides will be equal is if quantities inside the absolute value brackets are 1) equal or 2) equal but with opposite signs.

So |A| = B or -B*

1. Remove the absolute value bars

2. LHS = RHS or LHS = -RHS

3. Set up the two equations

Case 1: x - 2 = x + 3, OR

Case 2: x - 2 = -(x + 3)

4. Solve

Case 1:
x - 2 = x + 3
-2 = 3. Not a solution

CASE 2:
x - 2 = -(x + 3)
x - 2 = -x - 3
2x = -1
x = -$$\frac{1}{2}$$

5. Check x = -$$\frac{1}{2}$$ in equation. --> |-$$\frac{5}{2}$$| = |$$\frac{5}{2}$$|

Correct. x = -$$\frac{1}{2}$$. Answer B

Hope it helps

*and |B| = A or -A. Technically there are four cases. Case 3 is RHS = LHS. Case 4 is RHS = - LHS. These two cases are identical to Cases 1 and 2.
Case 3: +RHS = +LHS, (x + 3) = (x - 2) (= Case 1)
Case 4: +RHS = -LHS, (x + 3) = (- x + 2). Multiply both sides by (-1) --> (-x - 3) = (x - 2) (= Case 2)

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If |x - 2| = |x + 3|, x could equal   [#permalink] 30 Jul 2017, 10:16
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