+218 What values of x satisfy \(|x-4|+2|x+3|\le11\)? Express your answer in interval notation.
+359 To solve the inequality \( |x - 4| + 2|x + 3| \leq 11 \), we'll consider different cases depending on the sign of \( x - 4 \) and \( x + 3 \).
Case 1: \( x - 4 \geq 0 \) and \( x + 3 \geq 0 \)
In this case, both absolute values are positive, so we have:
\[ (x - 4) + 2(x + 3) \leq 11 \]
\[ x - 4 + 2x + 6 \leq 11 \]
\[ 3x + 2 \leq 11 \]
\[ 3x \leq 9 \]
\[ x \leq 3 \]
Case 2: \( x - 4 \geq 0 \) and \( x + 3 < 0 \)
In this case, \( x - 4 \geq 0 \) and \( x + 3 < 0 \), so \( x \) is between -3 and 4.
\[ (x - 4) + 2(-x - 3) \leq 11 \]
\[ x - 4 - 2x - 6 \leq 11 \]
\[ -x - 10 \leq 11 \]
\[ -x \leq 21 \]
\[ x \geq -21 \]
Case 3: \( x - 4 < 0 \) and \( x + 3 \geq 0 \)
In this case, \( x - 4 < 0 \) and \( x + 3 \geq 0 \), so \( x \) is between -3 and 4.
\[ -(x - 4) + 2(x + 3) \leq 11 \]
\[ -x + 4 + 2x + 6 \leq 11 \]
\[ x + 10 \leq 11 \]
\[ x \leq 1 \]
Case 4: \( x - 4 < 0 \) and \( x + 3 < 0 \)
In this case, both absolute values are negative, so we have:
\[ -(x - 4) + 2(-x - 3) \leq 11 \]
\[ -x + 4 - 2x - 6 \leq 11 \]
\[ -3x - 2 \leq 11 \]
\[ -3x \leq 13 \]
\[ x \geq -\frac{13}{3} \]
Now, let's combine the solutions from each case:
- Case 1: \( x \leq 3 \)
- Case 2: \( x \geq -21 \)
- Case 3: \( x \leq 1 \)
- Case 4: \( x \geq -\frac{13}{3} \)
The overlapping intervals are \( [-21, 3] \) and \( \left[ -\frac{13}{3}, 1 \right] \).
Therefore, the values of \( x \) satisfying the inequality are \( \boxed{[-21, 3] \cup \left[ -\frac{13}{3}, 1 \right]} \).
