If `|x + 4| + |x-4|=2|x| and |x+1|+|5-x|=6` then x belongs to

Options

A. [4, 6]
B. [-1,4]
C. [4, 5]
D. [-2, -1]

Answer

To solve the given equations step by step, we will analyze each equation separately and then find the common solution.

### Step 1: Solve the first equation \( |x + 4| + |x - 4| = 2|x| \)

**Case 1:** \( x \geq 4 \)

In this case:
\[ |x + 4| = x + 4 \quad \text{and} \quad |x - 4| = x - 4 \]
So the equation becomes:
\[ (x + 4) + (x - 4) = 2x \]
This simplifies to:
\[ 2x = 2x \]
This is always true, meaning there are infinitely many solutions for \( x \geq 4 \).

**Case 2:** \( 0 \leq x < 4 \)

In this case:
\[ |x + 4| = x + 4 \quad \text{and} \quad |x - 4| = 4 - x \]
So the equation becomes:
\[ (x + 4) + (4 - x) = 2x \]
This simplifies to:
\[ 8 = 2x \quad \Rightarrow \quad x = 4 \]
However, \( x = 4 \) is not in the interval \( 0 \leq x < 4 \), so there are no solutions in this case.

**Case 3:** \( -4 < x < 0 \)

In this case:
\[ |x + 4| = x + 4 \quad \text{and} \quad |x - 4| = 4 - x \]
So the equation becomes:
\[ (x + 4) + (4 - x) = -2x \]
This simplifies to:
\[ 8 = -2x \quad \Rightarrow \quad x = -4 \]
However, \( x = -4 \) is not in the interval \( -4 < x < 0 \), so there are no solutions in this case.

**Case 4:** \( x < -4 \)

In this case:
\[ |x + 4| = - (x + 4) \quad \text{and} \quad |x - 4| = - (x - 4) \]
So the equation becomes:
\[ -(x + 4) - (x - 4) = -2x \]
This simplifies to:
\[ -2x = -2x \]
This is always true, meaning there are infinitely many solutions for \( x < -4 \).

### Summary for the first equation:
The solutions for the first equation are:
\[ x \in (-\infty, -4) \cup [4, \infty) \]

### Step 2: Solve the second equation \( |x + 1| + |5 - x| = 6 \)

**Case 1:** \( x \geq 5 \)

In this case:
\[ |x + 1| = x + 1 \quad \text{and} \quad |5 - x| = -(5 - x) = x - 5 \]
So the equation becomes:
\[ (x + 1) + (x - 5) = 6 \]
This simplifies to:
\[ 2x - 4 = 6 \quad \Rightarrow \quad 2x = 10 \quad \Rightarrow \quad x = 5 \]
This is a valid solution.

**Case 2:** \( -1 \leq x < 5 \)

In this case:
\[ |x + 1| = x + 1 \quad \text{and} \quad |5 - x| = 5 - x \]
So the equation becomes:
\[ (x + 1) + (5 - x) = 6 \]
This simplifies to:
\[ 6 = 6 \]
This is always true, meaning there are infinitely many solutions for \( -1 \leq x < 5 \).

**Case 3:** \( x < -1 \)

In this case:
\[ |x + 1| = -(x + 1) \quad \text{and} \quad |5 - x| = 5 - x \]
So the equation becomes:
\[ -(x + 1) + (5 - x) = 6 \]
This simplifies to:
\[ 4 - 2x = 6 \quad \Rightarrow \quad -2x = 2 \quad \Rightarrow \quad x = -1 \]
However, \( x = -1 \) is not in the interval \( x < -1 \), so there are no solutions in this case.

### Summary for the second equation:
The solutions for the second equation are:
\[ x \in [-1, 5] \]

### Step 3: Find the common solution

Now we need to find the intersection of the solutions from both equations:
1. From the first equation: \( x \in (-\infty, -4) \cup [4, \infty) \)
2. From the second equation: \( x \in [-1, 5] \)

The common solution is:
\[ x \in [4, 5] \]

### Final Answer:
Thus, the final solution is:
\[ x \in [4, 5] \]