1. **Problem statement:** Given that $y = \frac{3x + 10}{x - 2}$, where $x$ and $y$ are positive and vary with time, and the rate of increase of $y$ is twice the rate of decrease of $x$, find the value of $y$. 2. **Step 1: Define variables and rates.** Let $\frac{dx}{dt}$ be the rate of change of $x$ with respect to time $t$, and $\frac{dy}{dt}$ be the rate of change of $y$ with respect to time. Given: $\frac{dy}{dt} = 2 \times \left(-\frac{dx}{dt}\right) = -2 \frac{dx}{dt}$ (since $x$ is decreasing, $\frac{dx}{dt} < 0$). 3. **Step 2: Differentiate $y$ with respect to $t$.** Using the quotient rule: $$y = \frac{3x + 10}{x - 2}$$ $$\frac{dy}{dt} = \frac{(3)(x - 2) - (3x + 10)(1)}{(x - 2)^2} \cdot \frac{dx}{dt}$$ Simplify numerator: $$3(x - 2) - (3x + 10) = 3x - 6 - 3x - 10 = -16$$ So, $$\frac{dy}{dt} = \frac{-16}{(x - 2)^2} \cdot \frac{dx}{dt}$$ 4. **Step 3: Use the given relation between rates:** $$\frac{dy}{dt} = -2 \frac{dx}{dt}$$ Substitute the expression for $\frac{dy}{dt}$: $$\frac{-16}{(x - 2)^2} \cdot \frac{dx}{dt} = -2 \frac{dx}{dt}$$ 5. **Step 4: Cancel $\frac{dx}{dt}$ (assuming $\frac{dx}{dt} \neq 0$):** $$\cancel{\frac{dx}{dt}} \cdot \frac{-16}{(x - 2)^2} = -2 \cancel{\frac{dx}{dt}}$$ This simplifies to: $$\frac{-16}{(x - 2)^2} = -2$$ 6. **Step 5: Solve for $x$:** Multiply both sides by $(x - 2)^2$: $$-16 = -2 (x - 2)^2$$ Divide both sides by $-2$: $$\frac{-16}{-2} = (x - 2)^2$$ $$8 = (x - 2)^2$$ Take square root: $$x - 2 = \pm \sqrt{8} = \pm 2\sqrt{2}$$ Since $x$ is positive and must satisfy $x - 2 > 0$ (denominator cannot be zero), choose positive root: $$x = 2 + 2\sqrt{2}$$ 7. **Step 6: Find $y$ at this $x$ value:** $$y = \frac{3x + 10}{x - 2} = \frac{3(2 + 2\sqrt{2}) + 10}{(2 + 2\sqrt{2}) - 2} = \frac{6 + 6\sqrt{2} + 10}{2\sqrt{2}} = \frac{16 + 6\sqrt{2}}{2\sqrt{2}}$$ 8. **Step 7: Simplify $y$:** Divide numerator and denominator: $$y = \frac{16}{2\sqrt{2}} + \frac{6\sqrt{2}}{2\sqrt{2}} = \frac{16}{2\sqrt{2}} + 3$$ Simplify first term: $$\frac{16}{2\sqrt{2}} = \frac{8}{\sqrt{2}} = 8 \cdot \frac{\sqrt{2}}{2} = 4\sqrt{2}$$ So, $$y = 4\sqrt{2} + 3$$ **Final answer:** $$\boxed{y = 4\sqrt{2} + 3}$$ This is the value of $y$ when the rate of increase of $y$ is twice the rate of decrease of $x$.