1. **State the problem:** Solve the equation $|x - a| + |x + a| = 4$ for $a$ given a fixed $x$. 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 $a = x$ and $a = -x$. 4. **Analyze the equation in three intervals for $a$:** - For $a \leq -x$: $|x - a| = x - a$, $|x + a| = -(x + a)$ So, $$|x - a| + |x + a| = (x - a) + (-(x + a)) = -2a$$ Set equal to 4: $$-2a = 4 \implies a = -2$$ Check if $a = -2 \leq -x$ to confirm validity. - For $-x < a < x$: $|x - a| = x - a$, $|x + a| = x + a$ So, $$|x - a| + |x + a| = (x - a) + (x + a) = 2x$$ Set equal to 4: $$2x = 4 \implies x = 2$$ This means for $x=2$, any $a$ in $(-2, 2)$ satisfies the equation. - For $a \geq x$: $|x - a| = a - x$, $|x + a| = x + a$ So, $$|x - a| + |x + a| = (a - x) + (x + a) = 2a$$ Set equal to 4: $$2a = 4 \implies a = 2$$ Check if $a = 2 \geq x$ to confirm validity. 5. **Summary of solutions:** - If $x < 2$, solutions are $a = -2$ (if $-2 \leq -x$) and $a = 2$ (if $2 \geq x$). - If $x = 2$, solution set is $a \in (-2, 2)$. 6. **Final answer:** - For $x = 2$, $a$ can be any value in $(-2, 2)$. - For $x < 2$, $a = -2$ if $-2 \leq -x$ and $a = 2$ if $2 \geq x$.