In this post, we will learn how to solve the given inequality \( |x – 3| + |x + 2| < 11 \). I will show a step-by-step approach to the solution.
Solution:
Using the definition of absolute value,
\( |x – 3| = \begin{cases} x – 3, & \text{if } x \geq 3, \\ -(x – 3), & \text{if } x < 3. \\ \end{cases} \)Similarly
\( |x + 2| = \begin{cases} x + 2, & \text{if } x \geq -2, \\ -(x + 2), & \text{if } x < -2. \\ \end{cases} \)These expressions show that the solution to \( |x – 3| + |x + 2| < 11 \) can be obtained by considering the following 3 cases:
1) Case 1: \( x < -2 \)
\( |x – 3| + |x + 2| < 11 \\ \Rightarrow (-x + 3) + (-x - 2) < 11 \\ \Rightarrow -2x + 1 < 11 \\ \Rightarrow -2x < 10 \\ \Rightarrow x > -5 \\ \)2) Case 2: \( -2 \leq x < 3 \)
\( |x – 3| + |x + 2| < 11 \\ \Rightarrow (-x + 3) + (x + 2) < 11 \\ \Rightarrow 5 < 11 \small \text{, (always true)} \\ \)3) Case 3: \( x \geq 3 \)
\( |x – 3| + |x + 2| < 11 \\ \Rightarrow (x - 3) + (x + 2) < 11 \\ \Rightarrow 2x < 12 \\ \Rightarrow x < 6 \\ \)Combining all three cases, we get \( -5 < x < 6 \). So, the solution to the given inequality \( |x – 3| + |x + 2| < 11 \) is
\[ \fbox{-5 < x < 6} \]
Please let me know in the comments if you find any error in this solution.