Differential equations not equivalent equations

So I have this problem which I partially understand (it's not solved by me, I am just trying to understand the solving), this is the equation: $$\frac{dx}{dt} = \frac{(2t + x + 2)}{(4t + 2x - 1)}$$ I circled the thing that I don't understand in the picture below. How are those two equations equivalent?

• $y-2t$ can been substituted in place of $x$, as $y$ has been defined to be $x+2t$ – GoodDeeds Feb 2 '16 at 14:10
• $\frac{d(y-2t)}{dt}=\frac{dy}{dt}-\frac{d(2t)}{dt}=\frac{dy}{dt}-2$ – GoodDeeds Feb 2 '16 at 14:10
• ohh okay, now I see, I didn't know that you can reduce $dt$, thanks for the explanation – southpaw93 Feb 2 '16 at 14:12
• @GoodDeeds you can submit and answer so I can vote it, if you want. – southpaw93 Feb 2 '16 at 14:14

$\frac{d(y-2t)}{dt}=\frac{dy}{dt}+\frac{d(-2t)}{dt}=\frac{dy}{dt}-2$
We have that $\dfrac{\partial (y-2t)}{\partial t}=\dfrac{\partial y}{\partial t}-2$
So,it's simple if you do this: $$\dfrac{\partial (y-2t)}{\partial t}=\dfrac{\partial y}{\partial t} + \dfrac{\partial (-2\cdot t)}{\partial t}$$
So, 2 is a clear number which is not depends on t.So $$\dfrac{\partial (y-2t)}{\partial t}=\dfrac{\partial y}{\partial t} + \dfrac{\partial (-2\cdot t)}{\partial t}=\dfrac{\partial y}{\partial t}-2\cdot \dfrac{\partial t}{\partial t}=\dfrac{\partial y}{\partial t}-2$$