|1-х|+|х+2|=3

Let's solve the equation: \[ |1 - x| + |x + 2| = 3 \] ### Step 1: Find critical points The absolute value expressions change behavior at points where their arguments are zero: - \(1 - x = 0 \Rightarrow x = 1\) - \(x + 2 = 0 \Rightarrow x = -2\) We will consider three intervals based on these points: 1. \(x < -2\) 2. \(-2 \leq x \leq 1\) 3. \(x > 1\) --- ### Step 2: Solve on each interval --- #### **Interval 1: \(x < -2\)** - \(1 - x > 0 \Rightarrow |1 - x| = 1 - x\) - \(x + 2 < 0 \Rightarrow |x + 2| = -(x + 2) = -x - 2\) Equation: \[ (1 - x) + (-x - 2) = 3 \] Simplify: \[ 1 - x - x - 2 = 3 \] \[ -2x - 1 = 3 \] \[ -2x = 4 \] \[ x = -2 \] Check if \(x = -2\) is in the interval \(x < -2\); **no**, it is at the boundary. --- #### **Interval 2: \(-2 \leq x \leq 1\)** - \(1 - x \geq 0 \Rightarrow |1 - x| = 1 - x\) - \(x + 2 \geq 0 \Rightarrow |x + 2| = x + 2\) Equation: \[ (1 - x) + (x + 2) = 3 \] Simplify: \[ 1 - x + x + 2 = 3 \] \[ 3 = 3 \] This simplifies to an identity, which means **every \(x\) in this interval satisfies the equation**. - So all \(x\) with \(-2 \leq x \leq 1\) are solutions. --- #### **Interval 3: \(x > 1\)** - \(1 - x < 0 \Rightarrow |1 - x| = x - 1\) - \(x + 2 > 0 \Rightarrow |x + 2| = x + 2\) Equation: \[ (x - 1) + (x + 2) = 3 \] Simplify: \[ x - 1 + x + 2 = 3 \] \[ 2x + 1 = 3 \] \[ 2x = 2 \] \[ x = 1 \] Check if \(x = 1\) belongs to this interval \(x > 1\); **no**, it is at the boundary. --- ### **Final answer:** The solution set includes all \(x\) in the interval \(-2 \leq x \leq 1\), plus the boundary points considered. **Answer:** \[ \boxed{ x \in [-2, 1] } \] This is the interval of solutions for the original equation.