
How do you write an inverse variation equation given $y = 1$ when $x = - 12$?
Answer
559.5k+ views
Hint: We have an equation $y = 1$ and we also know that $x = - 12$. In order to solve this particular equation, we should first be aware of the fact that for an inverse equation, it has always been represented by \[xy = k\]. This means that when one of the numbers exists, the value of the second variable varies inversely.
Complete step-by-step solution:
If x and y form an inverse variation then for this particular question, \[xy = k\] for some constant $k$, with the variables.
Given that,
$ \Rightarrow (x = - 12,y = - 1)\;$ is a solution to the required relation,
Now, putting the values inside the formula in order to know the value of the constant,
$ \Rightarrow - 12 \times - 1 = k$
Multiplying the given numbers to find the value of the constant,
$ \Rightarrow k = 12$
Now, we know the value of the constant, so putting it in the formula for the inverse variation equation.
That is, \[xy = k\]
$ \Rightarrow y = \dfrac{k}{x}$
Putting the values,
$ \Rightarrow y = \dfrac{{12}}{x}$
Explanation:
In an inverse variation - or inverse proportion, as one quantity increases the other decreases.
This can be written as: $y \propto \dfrac{1}{x}$
Variations (proportions) are linked by a constant $k$
We can write a variation as an equation by using the constant.
$ \Rightarrow y = \dfrac{k}{x}$
\[ \Rightarrow x \times y = k\]←now we can find a value for $k$ using the values of $x$ and $y$ which were given
$ \Rightarrow - 12 \times - 1 = k$
The equation is therefore: $y = \dfrac{{12}}{x}$.
Note: An inverse variation can be represented by the equation \[xy = k\] or $y = \dfrac{k}{x}$. That is, $y$ varies inversely as $x$ if there is some non-zero constant $k$ such that, \[xy = k\] or $y = \dfrac{k}{x}$ where $x \ne 0,y \ne 0$.
Suppose $y$ varies inversely as $x$ such that $xy = 3$ or $y = \dfrac{3}{x}$. While direct variation describes a linear relationship between two variables, inverse variation describes another kind of relationship. For two quantities with inverse variation, as one quantity increases, the other quantity decreases.
Complete step-by-step solution:
If x and y form an inverse variation then for this particular question, \[xy = k\] for some constant $k$, with the variables.
Given that,
$ \Rightarrow (x = - 12,y = - 1)\;$ is a solution to the required relation,
Now, putting the values inside the formula in order to know the value of the constant,
$ \Rightarrow - 12 \times - 1 = k$
Multiplying the given numbers to find the value of the constant,
$ \Rightarrow k = 12$
Now, we know the value of the constant, so putting it in the formula for the inverse variation equation.
That is, \[xy = k\]
$ \Rightarrow y = \dfrac{k}{x}$
Putting the values,
$ \Rightarrow y = \dfrac{{12}}{x}$
Explanation:
In an inverse variation - or inverse proportion, as one quantity increases the other decreases.
This can be written as: $y \propto \dfrac{1}{x}$
Variations (proportions) are linked by a constant $k$
We can write a variation as an equation by using the constant.
$ \Rightarrow y = \dfrac{k}{x}$
\[ \Rightarrow x \times y = k\]←now we can find a value for $k$ using the values of $x$ and $y$ which were given
$ \Rightarrow - 12 \times - 1 = k$
The equation is therefore: $y = \dfrac{{12}}{x}$.
Note: An inverse variation can be represented by the equation \[xy = k\] or $y = \dfrac{k}{x}$. That is, $y$ varies inversely as $x$ if there is some non-zero constant $k$ such that, \[xy = k\] or $y = \dfrac{k}{x}$ where $x \ne 0,y \ne 0$.
Suppose $y$ varies inversely as $x$ such that $xy = 3$ or $y = \dfrac{3}{x}$. While direct variation describes a linear relationship between two variables, inverse variation describes another kind of relationship. For two quantities with inverse variation, as one quantity increases, the other quantity decreases.
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