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If $y$ varies directly with $x$ and $y=4$ when $x=12$, how do you find $x$ when $y=36$?

Answer
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Hint: For these kinds of questions, we need to make use of the small concept of differential equations. We are told that $y$ varies directly with $x$ . But not $y$varies the same as the variation in $x$. We should be able to infer that the rate of change of $y$is proportional to the rate of change of $x$ but not equal to the rate of change of $x$. So let us write down an equation to mathematically show that and use integration to find out the exact equation relating both $x,y.$

Complete step by step answer:
The rate of change of $y$ which can be represented as $\dfrac{dy}{dx}$ is proportional to the rate of change of $x$ which can be represented as $\dfrac{dx}{dx}$ which is nothing but $1$.
Here, we should keep in mind that we are differentiating with respect to $x$.
So now let's put down the equation which represents that the rate of change of $y$ is proportional to the rate of change of $x$.
It is as follows :
$\Rightarrow \dfrac{dy}{dx}$ $\propto $$1$
Now, we all know that when we try to remove the proportionality symbol, a constant is to be multiplied since the variables always maintain a constant ratio to each other.
So, now let us remove the proportionality symbol and multiply any constant, let us say $k$.
Upon doing so, we get the following :
$\Rightarrow \dfrac{dy}{dx}$$\propto $$1$
\[\Rightarrow \dfrac{dy}{dx}=k\]
Now let us integrate on both sides.
Upon integrating on both sides, we get the following :
$\Rightarrow \dfrac{dy}{dx}$$\alpha $$1$
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=k \\
 & \Rightarrow \int{\dfrac{dy}{dx}=\int{k}} \\
 & \Rightarrow y=kx \\
\end{align}\]
From the condition, we have a base condition which is $y=-4.5$ when $x=13.5$.
So let us use this piece of information to find the value of $k$.
$\begin{align}
  & \Rightarrow y=kx \\
 & \Rightarrow 4=12k \\
 & \Rightarrow 1=3k \\
 & \Rightarrow k=\dfrac{1}{3} \\
\end{align}$
So the exact equation relating both $x,y$ is $y=\dfrac{x}{3}$.
Now, let us find out the value of $x$ when the value of $y=36$.
Let us plug in $y=36$ in the equation $y=\dfrac{x}{3}$to find $x$.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow y=\dfrac{x}{3} \\
 & \Rightarrow 36=\dfrac{x}{3} \\
 & \Rightarrow x=108 \\
\end{align}$

$\therefore $ Hence, the value of $x$ when $y=36$ is $108$.

Note: After we integrated our differential expression, we didn’t add an extra constant, $c$, which we usually do. If we did add that constant as well, then we would have two unknown variables in our expression namely $k,c$ but we have only one condition. So we will not get the absolute values of either of the constants. We would only get a relation between them. If that happens, we can’t form an exact equation between $x,y$ and fail to find out the value $x$ when $y=36$.
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