1. The problem is to solve the equation $x^2 - y^2 = 3$ for $y$ in terms of $x$. 2. Start by isolating $y^2$ on one side: $$x^2 - y^2 = 3 \implies y^2 = x^2 - 3$$ 3. To solve for $y$, take the square root of both sides: $$y = \pm \sqrt{x^2 - 3}$$ 4. Note that the expression under the square root, $x^2 - 3$, must be non-negative for real values of $y$: $$x^2 - 3 \geq 0 \implies x^2 \geq 3 \implies |x| \geq \sqrt{3}$$ 5. Therefore, the solution set for $y$ is: $$y = \pm \sqrt{x^2 - 3} \quad \text{for} \quad |x| \geq \sqrt{3}$$ This means $y$ is real only when $x$ is less than or equal to $-\sqrt{3}$ or greater than or equal to $\sqrt{3}$. Final answer: $$y = \pm \sqrt{x^2 - 3}$$ with domain $|x| \geq \sqrt{3}$.