Question

If $y$ varies directly with $x$ and $y=-4.5$ when $x=13.5$ , how do you find $x$ when $y=24$ ?

Answer 1

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 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 solution:
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 it 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 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, 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 :
$\Rightarrow \dfrac{dy}{dx}$ $\alpha$ $1$

& \Rightarrow \dfrac{dy}{dx}=k \\\ & \Rightarrow \int{\dfrac{dy}{dx}=\int{k}} \\\ & \Rightarrow y=kx \\\ \end{aligned}$$ From the condition, we have a base condition which is $ y=13.5 $ when $ x=-4.5 $ . So let us use this piece of information to find the value of $ k $ . $ \begin{aligned} & \Rightarrow y=kx \\\ & \Rightarrow -4.5=13.5k \\\ & \Rightarrow -45=135k \\\ & \Rightarrow k=\dfrac{-1}{3} \\\ \end{aligned} $ 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=24 $ . Let us plug in $ y=24 $ in the equation $ y=\dfrac{-x}{3} $ to find $ x $ . Upon doing so, we get the following : $ \begin{aligned} & \Rightarrow y=\dfrac{-x}{3} \\\ & \Rightarrow 24=\dfrac{-x}{3} \\\ & \Rightarrow x=-72 \\\ \end{aligned} $ $ \therefore $ Hence, the value of $ x $ when $ y=24 $ is $ -72 $ . **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=24 $ .