# If $y \propto \dfrac{1}{x}$ and $x = 7$ when $y = 9$, find the constant of variation $\left( k \right)$.

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Hint- Here, we will be putting the given values of the variables in the equation obtained after removing the inversely proportional sign.

Given $y \propto \dfrac{1}{x}$

Let the constant of variation introduced be $k$ after removing the proportionality sign.

\[y \propto \dfrac{1}{x} \Rightarrow y = k\left( {\dfrac{1}{x}} \right){\text{ }} \to {\text{(1)}}\]

Also given that when $y = 9$, the corresponding value of $y$ is $y = 9$.

Put $x = 7$ and $y = 9$ in equation (1), we get

\[ \Rightarrow y = k\left( {\dfrac{1}{x}} \right){\text{ }} \Rightarrow 9 = k\left( {\dfrac{1}{7}} \right) \Rightarrow k = 7 \times 9 = 63\]

Therefore, the required value of the constant of variation $\left( k \right)$ is 63.

Note- These types of problems are solved by assuming a constant of proportionality or a constant of variation when the proportionality sign (either directly proportional or inversely proportional) is removed.

Given $y \propto \dfrac{1}{x}$

Let the constant of variation introduced be $k$ after removing the proportionality sign.

\[y \propto \dfrac{1}{x} \Rightarrow y = k\left( {\dfrac{1}{x}} \right){\text{ }} \to {\text{(1)}}\]

Also given that when $y = 9$, the corresponding value of $y$ is $y = 9$.

Put $x = 7$ and $y = 9$ in equation (1), we get

\[ \Rightarrow y = k\left( {\dfrac{1}{x}} \right){\text{ }} \Rightarrow 9 = k\left( {\dfrac{1}{7}} \right) \Rightarrow k = 7 \times 9 = 63\]

Therefore, the required value of the constant of variation $\left( k \right)$ is 63.

Note- These types of problems are solved by assuming a constant of proportionality or a constant of variation when the proportionality sign (either directly proportional or inversely proportional) is removed.

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