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# Is the equation $6{x^2} + 3y = 0$an example of direct variation?

Last updated date: 16th Sep 2024
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Hint: We know that two variables are in direct variation when the value of one is always a fixed multiple of the other. If we are able to prove that, then they are in direct variation. Otherwise, they will be following some other relation such as inverse variation.

In the above question, it is given that $6{x^2} + 3y = 0$.
we have to find whether it follows direct variation or not. But for that first we have to understand what is a direct variation.
We can say that two variables are in direct variation when the value of one is always a fixed multiple of the other. This means, for example, when one of the variables doubles, the other must also double.
for variables are x and y, and y is always k times as much as x, then x and y are in direct variation. This relation is written as:
$y = k \times x$
Here's an example: picture a fence. That fence's length can be measured in yards, but it can also be measured in feet.

Now, let's double the length of the fence.
As a result, the number of yards in the fence has doubled, but so has the number of feet. This means the fence's length in feet is in direct variation with (or is directly proportional to) its length in yards.

This relation is quite easy to deduce. Because there are 3 feet in a yard, we have
$[numberof{\text{ }}feet] = 3 \times [numberof{\text{ }}yards]$
$y = 3 \times x$
This equation could also be written as $\dfrac{y}{x} = 3$, to emphasize that the ratio between y and x is a constant $3$.
This is different than an equation of the form $\dfrac{k}{x} = y$; here, when x gets doubled, y gets halved. In this case, x and y are said to be inversely proportion.
Now, you were given the equation $6{x^2} + 3y = 0\;$and asked whether this was a direct variation.
$6{x^2} + 3y = 0\;$
Now subtract $6{x^2}$from both sides
$3y = 0 - 6{x^2}$
$3y = - 6{x^2}$
Now, divide both sides by $3$
$y = \dfrac{{ - 6{x^2}}}{3}$
$y = - 2{x^2}$
There we go, we can see that y is directly proportional to ${x^2}$, but not to x itself.
Example: when $x = 3$, we have $y = - 2{(3)^2} = - 2(9) = - 18$.
If we double x to $6$, we get $y = - 2{(6)^2} = - 2(36) = - 72$, which is $4 \times - 18$.
Therefore, we can see that when x got doubled, y got quadrupled.
Hence, this is not an example of direct variation.

Note: If a change in the value of a variable results in a change in the value of a related variable, their relationship is termed as a variation. If the value of a variable increases with an increase in the value of a related variable, their relationship is termed as a direct variation. If the value of a variable decreases with an increase in the value of a related variable or vice versa, their relationship is termed as an inverse variation.