1. **State the problem:** Given the function $y = \frac{3x + 10}{x - 2}$ 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. **Write down the given function:** $$y = \frac{3x + 10}{x - 2}$$ 3. **Differentiate $y$ with respect to time $t$ using the quotient rule:** The quotient rule states: $$\frac{dy}{dt} = \frac{(v)(du/dt) - (u)(dv/dt)}{v^2}$$ where $u = 3x + 10$ and $v = x - 2$. Calculate derivatives: $$\frac{du}{dt} = 3 \frac{dx}{dt}$$ $$\frac{dv}{dt} = \frac{dx}{dt}$$ Apply quotient rule: $$\frac{dy}{dt} = \frac{(x - 2)(3 \frac{dx}{dt}) - (3x + 10)(\frac{dx}{dt})}{(x - 2)^2}$$ 4. **Simplify numerator:** $$= \frac{3(x - 2) \frac{dx}{dt} - (3x + 10) \frac{dx}{dt}}{(x - 2)^2}$$ $$= \frac{\frac{dx}{dt} [3(x - 2) - (3x + 10)]}{(x - 2)^2}$$ Simplify inside brackets: $$3(x - 2) - (3x + 10) = 3x - 6 - 3x - 10 = -16$$ So: $$\frac{dy}{dt} = \frac{\frac{dx}{dt} (-16)}{(x - 2)^2} = -16 \frac{\frac{dx}{dt}}{(x - 2)^2}$$ 5. **Use the given relation between rates:** The rate of increase of $y$ is twice the rate of decrease of $x$. Since $x$ is decreasing, $\frac{dx}{dt} < 0$. Let $\frac{dx}{dt} = -k$ where $k > 0$. Then: $$\frac{dy}{dt} = 2k$$ From step 4: $$\frac{dy}{dt} = -16 \frac{\frac{dx}{dt}}{(x - 2)^2} = -16 \frac{-k}{(x - 2)^2} = \frac{16k}{(x - 2)^2}$$ Set equal to $2k$: $$\frac{16k}{(x - 2)^2} = 2k$$ 6. **Solve for $x$:** Divide both sides by $k$ (nonzero): $$\frac{16}{(x - 2)^2} = 2$$ Multiply both sides by $(x - 2)^2$: $$16 = 2 (x - 2)^2$$ Divide both sides by 2: $$\cancel{16}^\cancel{} = \cancel{2} (x - 2)^2 / \cancel{2}$$ $$8 = (x - 2)^2$$ Take square root: $$x - 2 = \pm \sqrt{8} = \pm 2\sqrt{2}$$ Since $x$ is positive and $x - 2$ must be positive (denominator positive), choose positive root: $$x - 2 = 2\sqrt{2}$$ $$x = 2 + 2\sqrt{2}$$ 7. **Find $y$ at this $x$ value:** $$y = \frac{3x + 10}{x - 2} = \frac{3(2 + 2\sqrt{2}) + 10}{2 + 2\sqrt{2} - 2} = \frac{6 + 6\sqrt{2} + 10}{2\sqrt{2}} = \frac{16 + 6\sqrt{2}}{2\sqrt{2}}$$ Simplify numerator and denominator: $$= \frac{16}{2\sqrt{2}} + \frac{6\sqrt{2}}{2\sqrt{2}} = \frac{16}{2\sqrt{2}} + 3$$ Simplify first term: $$\frac{16}{2\sqrt{2}} = \frac{8}{\sqrt{2}} = 8 \frac{\sqrt{2}}{2} = 4\sqrt{2}$$ So: $$y = 4\sqrt{2} + 3$$ **Final answer:** $$\boxed{y = 4\sqrt{2} + 3}$$