If x is an integer, then how many values of x will satisfy the equatio

Question

If x is an integer, then how many values of x will satisfy the equation |x + 2|= |3x +14|?

A. 0
B. 1
C. 2
D. 3
E. 4

Different methods to solve absolute value equations and inequalities- Exercise Question #3

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jaysonbeatty12 · Accepted Answer

Absolute value is the easiest / hardest concept. The GMAT uses absolute value as a proxy for logical reasoning. Students frequently get fooled by absolute value questions, so my recommendation is whenever possible rewrite the equation without absolute value. This investment of time, in my experience, greatly increases accuracy on these questions.

In this case, if we get rid of absolute value we are left with two equations:

1. x + 2 = 3x + 14
2. x + 2 = -(3x + 14)

Please note that the two other possibilities of -(x + 2) = 3x + 14 and -(x + 2) = -(3x + 14) reduce to either 1 or 2.

Checking in the with the answer choices, only B and C are now alive. Either 1 and 2 have unique values or both solve for the same value. The latter is less likely in real life but could a devious testmaker trick, so you have to check.

1. becomes -2x = 12 then x = -6
2. becomes x + 2 = -3x - 14 then 4x = -16 then x = -4

C is correct

CounterSniper · Answer

if 
|x| = |y|
then 
 x^2  =  y^2 
similarly 
 (x + 2)^2 =  (3x +14)^2 
upon solving ,
we get x= -6 and -4

C

EgmatQuantExpert · Answer

Solution 
  Given:  
 • We are given that x is an integer, and
• We are also given an absolute value equation, |x + 2| = |3x + 14|

To find:  
 • We need to find the number of values of x, that satisfy the given equation

Approach and Working:   
 • In this question, we have modulus function on both sides of the equation.
• Let’s first apply the definition to the LHS of the equation.
• From this, we get,
 o x + 2 = |3x + 14|, if x ≥ -2, and
o x + 2 = -|3x + 14|, if x < -2 
• Now, let’s consider the first case, x + 2 = |3x + 14|, if x ≥ -2. And, if we remove the modulus sign on the RHS of this equation, we get,
 o x + 2 = 3x + 14, if x ≥ -2 and x ≥ -14/3
  Solving this equation, we get, 2x = -12
 Implies, x = -6, which does not lie in the range
 Thus, x = -6 is not a possible value 
o x + 2 = -(3x + 14), if x ≥ -2 and x < -14/3
  This case is not possible as no value of x can simultaneously be ≥ -2 and < -14/3  
• Considering the second case, x + 2 = -|3x + 14|, if x < -2, and removing the modulus sign, we get,
 o x + 2 = -(3x + 14), if x < -2 and x ≥ -14/3
  Solving this equation, we get, 4x = -16
 Implies, x = -4, which lies in the range  
o x + 2 = (3x + 14), if x < -2 and x < -14/3
  Solving this equation, we get, 2x = -12
 Implies, x = -6, which is less than both -2 and -14/3
 Thus, x = -6 is a possible value of x

Therefore, the only values of x, that satisfy the given equation, |x + 2|= |3x +14|, are -4 and -6

Hence, the correct answer is option C.

Answer: C

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TariqOmar · Answer

Correct answer is B.

Solving the equation will yield two solutions , X=-6 and X=-3. Plugging these values into the equation will yield only 1 value to be valid, and that value is -6.

ssgodfather · Answer

But then don't you think the answer can be both 6 as well as -6? Since the equation is the mod of x? So I think the answer should be c!

sayankh · Answer

'C' it is, squaring both sides.

x^2 + 4x + 4 = 9x^2 + 84x +196

8x^2 + 80x + 192 = 0

x^2 + 10x + 24 = 0

(x+6)(x+4)=0

x = -4, -6

HUNTdGMAT · Answer

You made me remember my old school days

anirchat · Answer

Isn't it correct that any equations of the type |ax+c| = |bx+d|, will have 2 solutions ?  EgmatQuantExpert

bumpbot · Answer

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If x is an integer, then how many values of x will satisfy the equatio

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