1. **State the problem:** Solve the equation $|x - a| + |x + a| = 4$ for $x$. 2. **Recall the definition of absolute value:** - $|y| = y$ if $y \geq 0$ - $|y| = -y$ if $y < 0$ 3. **Consider the critical points where the expressions inside the absolute values change sign:** These are at $x = a$ and $x = -a$. 4. **Analyze the equation in three intervals:** - For $x \geq a$: $$|x - a| = x - a, \quad |x + a| = x + a$$ So, $$|x - a| + |x + a| = (x - a) + (x + a) = 2x$$ Set equal to 4: $$2x = 4 \implies x = 2$$ Check if $x=2 \geq a$ to confirm validity. - For $-a \leq x < a$: $$|x - a| = a - x, \quad |x + a| = x + a$$ So, $$|x - a| + |x + a| = (a - x) + (x + a) = 2a$$ Set equal to 4: $$2a = 4 \implies a = 2$$ This means for $a=2$, any $x$ in $[-2,2]$ satisfies the equation. - For $x < -a$: $$|x - a| = a - x, \quad |x + a| = -x - a$$ So, $$|x - a| + |x + a| = (a - x) + (-x - a) = -2x$$ Set equal to 4: $$-2x = 4 \implies x = -2$$ Check if $x = -2 < -a$ to confirm validity. 5. **Summary of solutions:** - If $a < 2$, solutions are $x = 2$ (if $2 \geq a$) and $x = -2$ (if $-2 < -a$), but $-2 < -a$ means $a < 2$ which is true. - If $a = 2$, all $x$ in $[-2,2]$ satisfy the equation. 6. **Final answer:** - For $a = 2$, solution set is $x \in [-2, 2]$. - For $a < 2$, solutions are $x = 2$ and $x = -2$ if they satisfy the interval conditions.