# Is 3y + 2= 2x a direct variation equation and if so, what is the constant of variation?

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Alan P. Share
Jan 26, 2016

$3 y + 2 = 2 x$ is not a direct variation equation

#### Explanation:

Method 1 For Seeing This
A direct variation equation will always pass through the origin
i.e. $\left(x , y\right) = \left(0 , 0\right)$ will always be a solution to the equation.
$\textcolor{w h i t e}{\text{XXX}}$for $3 y + 2 = 2 x$
$\textcolor{w h i t e}{\text{XXX}} 3 \left(0\right) + 2 \ne 2 \left(0\right)$ so this condition is not satisfied.

Method 2 For Seeing This
For a direct variation equation, if $\left(a , b\right)$ is a solution, then $\left(c \times a , c \times b\right)$ is also a solution (for any constant $c$).
$\textcolor{w h i t e}{\text{XXX}}$Noting that $\left(4 , 3\right)$ is a solution to $3 y + 2 = 2 x$
$\textcolor{w h i t e}{\text{XXX}}$We can check using $c = 5$
$\textcolor{w h i t e}{\text{XXX}}$to se if $\left(x , y\right) = \left(5 \times 4 , 5 \times 3\right) = \left(20 , 15\right)$ is a solution.
$\textcolor{w h i t e}{\text{XXX}} 3 \left(15\right) + 2 = 47 \ne 40 = 2 \left(20\right)$
$\textcolor{w h i t e}{\text{XXXXXX}}$So, once again, we see that this is not a direct variation.

Method 3 For Seeing This
Any direct variation can be transformed into the form:
$\textcolor{w h i t e}{\text{XXX}} y = m \cdot x$ for some constant $m$
$\textcolor{w h i t e}{\text{XXX}} 3 y + 2 = 2 x$ can be transformed into
$\textcolor{w h i t e}{\text{XXXXXX}} y = \frac{2}{3} x - \frac{2}{3}$
$\textcolor{w h i t e}{\text{XXX}}$but there is no way to dispose of the $\left(- \frac{2}{3}\right)$ to get it into the above form.

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