Given that |12 − 4x| + 2 = 6 and we need to find which value of x out of the option choices satisfies this.

Let's understand how to solve this problem using two methods

Method 1: Substitution

Let's take each answer choice and substitute in the question and check which one satisfies the question

A. −4 Put x = -4 in |12 − 4x| + 2 = 6. We get
| 12 - 4*(-4)| + 2 = | 12 + 16| + 2 = |28| + 2 = 28 + 2 = 30 \(\neq\) 6 => NOT POSSIBLE

B. −2 Put x = -2 in |12 − 4x| + 2 = 6. We get
| 12 - 4*(-2)| + 2 = | 12 + 8| + 2 = |20| + 2 = 20 + 2 = 22 \(\neq\) 6 => NOT POSSIBLE

C. 1 Put x = 1 in |12 − 4x| + 2 = 6. We get
| 12 - 4*1| + 2 = | 12 - 4 | + 2 = |8| + 2 = 8 + 2 = 10 \(\neq\) 6 => NOT POSSIBLE

D. 3 Put x = 3 in |12 − 4x| + 2 = 6. We get
| 12 - 4*3| + 2 = | 12 - 12| + 2 = |0| + 2 = 0 + 2 = 2 \(\neq\) 6 => NOT POSSIBLE

E. 4 Put x = 4 in |12 − 4x| + 2 = 6. We get
| 12 - 4*4| + 2 = | 12 - 16| + 2 = |-4| + 2 = 4 + 2 = 6 = 6 => POSSIBLE

Method 2: Algebra

|12 − 4x| + 2 = 6
=> |12 − 4x| = 6 -2 = 4
Divide both the sides by 4 we get
\(\frac{|12-4x|}{4} = \frac{4}{4}\)
=> | 3 - x | = 1

Now, there are two ways of solving this

Method 2.1: Squaring

Square both the sides we get
\((|3-x|)^2 = 1^2\)
=> \((3-x) ^ 2 \)= 1
=> \(3^2 + x^2 - 2*3*x\) = 1
=> 9 + \(x^2\) - 6x - 1 = 0
=> \(x^2\) - 6x + 8 = 0
=> \(x^2\) - 2x - 4x + 8 =0
=> x*(x-2) -4*(x-2) = 0
=> (x-2) * (x-4) = 0
=> x = 2 or x = 4

Method 2.2: Opening Absolute Value

| 3 - x | = 1
=> 3-x = 1 or 3-x =-1
=> x = 3-1 = 2 or x = 3+1 = 4

So, Answer will be E
Hope it helps!

Watch the following video to MASTER Absolute Values