5 It is given that $$ y=\frac{3 x+10}{x-2} $$, where $$ x $$ and $$ y $$ are positive and vary with time. Given that the rate of increase of $$ y $$ is twice the rate of decrease of $$ x $$, find the value of $$ y $$.

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To solve the problem, we start with the given equation:

$$
y = \frac{3x + 10}{x - 2}
$$

We need to find the relationship between the rates of change of $$ y $$ and $$ x $$. The problem states that the rate of increase of $$ y $$ is twice the rate of decrease of $$ x $$.

Let's denote:
- $$ \frac{dy}{dt} $$ as the rate of change of $$ y $$ with respect to time $$ t $$
- $$ \frac{dx}{dt} $$ as the rate of change of $$ x $$ with respect to time $$ t $$

From the problem, we have:

$$
\frac{dy}{dt} = -2 \frac{dx}{dt}
$$

Next, we will differentiate $$ y $$ with respect to $$ t $$ using the chain rule. The derivative of $$ y $$ with respect to $$ x $$ is:

$$
\frac{dy}{dx} = \frac{(x - 2)(3) - (3x + 10)(1)}{(x - 2)^2}
$$

Now, let's calculate $$ \frac{dy}{dx} $$:
Simplify the expression by following steps:
- step0: Solution:
  $$\frac{\left(x-2\right)\times 3-\left(3x+10\right)\times 1}{\left(x-2\right)^{2}}$$
- step1: Multiply the terms:
  $$\frac{3\left(x-2\right)-\left(3x+10\right)\times 1}{\left(x-2\right)^{2}}$$
- step2: Multiply:
  $$\frac{3\left(x-2\right)-3x-10}{\left(x-2\right)^{2}}$$
- step3: Subtract the terms:
  $$\frac{-16}{\left(x-2\right)^{2}}$$
- step4: Rewrite the fraction:
  $$-\frac{16}{\left(x-2\right)^{2}}$$
- step5: Calculate:
  $$-\frac{16}{x^{2}-4x+4}$$
The derivative of $$ y $$ with respect to $$ x $$ is:

$$
\frac{dy}{dx} = -\frac{16}{(x - 2)^2}
$$

Now, we can express $$ \frac{dy}{dt} $$ in terms of $$ \frac{dx}{dt} $$:

$$
\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = -\frac{16}{(x - 2)^2} \cdot \frac{dx}{dt}
$$

Substituting this into the equation $$ \frac{dy}{dt} = -2 \frac{dx}{dt} $$:

$$
-\frac{16}{(x - 2)^2} \cdot \frac{dx}{dt} = -2 \frac{dx}{dt}
$$

Assuming $$ \frac{dx}{dt} \neq 0 $$, we can divide both sides by $$ \frac{dx}{dt} $$:

$$
-\frac{16}{(x - 2)^2} = -2
$$

Now, we can simplify this equation:

$$
\frac{16}{(x - 2)^2} = 2
$$

Multiplying both sides by $$ (x - 2)^2 $$:

$$
16 = 2(x - 2)^2
$$

Dividing both sides by 2:

$$
8 = (x - 2)^2
$$

Taking the square root of both sides:

$$
x - 2 = \pm \sqrt{8}
$$

Since $$ x $$ is positive, we take the positive root:

$$
x - 2 = 2\sqrt{2} \implies x = 2 + 2\sqrt{2}
$$

Now, we can substitute this value of $$ x $$ back into the equation for $$ y $$:

$$
y = \frac{3(2 + 2\sqrt{2}) + 10}{(2 + 2\sqrt{2}) - 2}
$$

Let's calculate $$ y $$:
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{3\left(2+2\sqrt{2}\right)+10}{\left(2+2\sqrt{2}\right)-2}$$
- step1: Remove the parentheses:
  $$\frac{3\left(2+2\sqrt{2}\right)+10}{2+2\sqrt{2}-2}$$
- step2: Multiply the numbers:
  $$\frac{6+6\sqrt{2}+10}{2+2\sqrt{2}-2}$$
- step3: Add the numbers:
  $$\frac{16+6\sqrt{2}}{2+2\sqrt{2}-2}$$
- step4: Remove the opposites:
  $$\frac{16+6\sqrt{2}}{2\sqrt{2}}$$
- step5: Rewrite the expression:
  $$\frac{2\left(8+3\sqrt{2}\right)}{2\sqrt{2}}$$
- step6: Reduce the fraction:
  $$\frac{8+3\sqrt{2}}{\sqrt{2}}$$
- step7: Multiply by the Conjugate:
  $$\frac{\left(8+3\sqrt{2}\right)\sqrt{2}}{\sqrt{2}\times \sqrt{2}}$$
- step8: Multiply the numbers:
  $$\frac{8\sqrt{2}+6}{\sqrt{2}\times \sqrt{2}}$$
- step9: Multiply the numbers:
  $$\frac{8\sqrt{2}+6}{2}$$
- step10: Rewrite the expression:
  $$\frac{2\left(4\sqrt{2}+3\right)}{2}$$
- step11: Reduce the fraction:
  $$4\sqrt{2}+3$$
The value of $$ y $$ is:

$$
y = 4\sqrt{2} + 3
$$

This is the final answer.
The value of $$ y $$ is $$ 4\sqrt{2} + 3 $$.

5 It is given that , where
and are positive and vary with time.
Given that the rate of increase of is
twice the rate of decrease of , find
the value of .

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5 It is given that , where and are positive and vary with time. Given that the rate of increase of is twice the rate of decrease of , find the value of .