1. **State the problem:** Solve the equation $3x^2 + 2y^2 = 35$ for one variable or analyze its properties. 2. **Identify the equation type:** This is an equation of an ellipse in terms of $x$ and $y$. 3. **Rewrite the equation:** $$3x^2 + 2y^2 = 35$$ 4. **Isolate $y^2$ to express $y$ in terms of $x$:** $$2y^2 = 35 - 3x^2$$ $$y^2 = \frac{35 - 3x^2}{2}$$ 5. **Express $y$ explicitly:** $$y = \pm \sqrt{\frac{35 - 3x^2}{2}}$$ 6. **Domain restriction:** For $y$ to be real, the expression under the square root must be non-negative: $$35 - 3x^2 \geq 0$$ $$3x^2 \leq 35$$ $$x^2 \leq \frac{35}{3}$$ $$-\sqrt{\frac{35}{3}} \leq x \leq \sqrt{\frac{35}{3}}$$ 7. **Summary:** The equation represents an ellipse centered at the origin with the above domain for $x$ and corresponding $y$ values. **Final answer:** $$y = \pm \sqrt{\frac{35 - 3x^2}{2}}$$