Question

If $y$ varies inversely as $x$, and \[y = \dfrac{2}{5}\] when $x = 2$, how do you find $y$ when $x = 1$?

Answer 1

Hint:The statement: y varies inversely as x means that when x increases, y decreases by the same factor. In other words, the expression \[xy\] is constant, to solve the inverse variation questions, we must know the equation \[y = \dfrac{k}{x}\], in which the we need to find the value of $k$ by \[k = yx\], considering the values of $x$ and $y$, then find $y$ when $x = 1$.

Formula used:
\[y = \dfrac{k}{x}\]
In which, $k$ is the constant of variation and $x$ and $y$ are the variables.

Complete step by step answer:
 Let us write the given data:
$y$ varies inversely as $x$,
\[y = \dfrac{2}{5}\] when \[x = 2\]
Hence, we need to find:
$y$ =? when $x=1$
We know that the inverse variation sums are solved using the equation:
\[y\alpha k \times \dfrac{1}{x}\]
\[ \Rightarrow \]\[y = \dfrac{k}{x}\]
Given condition that y varies inversely with x, and \[y = \dfrac{2}{5}\] when \[x = 2\]i.e.,
\[\left( {x,y} \right) \to \left( {2,\dfrac{2}{5}} \right)\]
Hence, by the given condition we have:
\[y = \dfrac{k}{x}\]
\[ \Rightarrow \]\[k = yx\]
Substitute the value of x and y as:
\[k = \dfrac{2}{5} \times 2\]
\[ \Rightarrow k = \dfrac{4}{5}\]
Hence, the equation is
\[y = \dfrac{4}{{5x}}\]
As, it is mentioned in the question that $x = 1$, then we get
\[y = \dfrac{4}{{5 \times 1}}\]
\[\therefore y = \dfrac{4}{5}\]

Therefore, the value of y when x = 1 is \[y = \dfrac{4}{5}\].

Note:Inverse variation equation are solved using the equation \[y = \dfrac{k}{x}\], here, y varies inversely as x with the given values we need to find the value of y with the condition we need to find the value of k with the equation as \[k = yx\], to determine if there are any other changes in the inverse variation equation. Use the information given in the problem to find the value of k, called the constant of variation or the constant of proportionality.