1. **State the problem:** Solve the equation $$\sqrt{y} + 5 \times \sqrt{x} - 3 = 4$$ for $y$ in terms of $x$. 2. **Rewrite the equation:** Add 3 to both sides to isolate the square root terms: $$\sqrt{y} + 5 \sqrt{x} = 7$$ 3. **Isolate $\sqrt{y}$:** $$\sqrt{y} = 7 - 5 \sqrt{x}$$ 4. **Square both sides** to eliminate the square root on $y$: $$y = (7 - 5 \sqrt{x})^2$$ 5. **Expand the right side:** $$y = 7^2 - 2 \times 7 \times 5 \sqrt{x} + (5 \sqrt{x})^2 = 49 - 70 \sqrt{x} + 25x$$ 6. **Final expression:** $$y = 49 - 70 \sqrt{x} + 25x$$ 7. **Domain considerations:** Since $\sqrt{x}$ and $\sqrt{y}$ are defined only for $x \geq 0$ and $y \geq 0$, and $\sqrt{y} = 7 - 5 \sqrt{x} \geq 0$, we have: $$7 - 5 \sqrt{x} \geq 0 \implies \sqrt{x} \leq \frac{7}{5} \implies x \leq \left(\frac{7}{5}\right)^2 = \frac{49}{25}$$ Thus, the domain is: $$0 \leq x \leq \frac{49}{25}$$ 8. **Range:** Since $y = (7 - 5 \sqrt{x})^2$, as $x$ goes from $0$ to $\frac{49}{25}$, $y$ goes from: $$y(0) = 7^2 = 49$$ $$y\left(\frac{49}{25}\right) = 0$$ So the range is: $$0 \leq y \leq 49$$