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**HallsofIvy** Your original equation is $\displaystyle \frac{dy}{dx}= \frac{4x- y+ 7}{2x+ y- 1}$. You make the substitution x= X- 1 (NOT "x= X+ 1"), y= Y+ 3. $\displaystyle \frac{dy}{dx}= \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}= \frac{dY}{dX}$. 4x-y+ 7= 4(X- 1)- Y- 3+ 7= 4X- Y and 2x+ y- 1= 2(X- 1)+ Y+ 3- 1= 2X+ Y. Yes, that makes the equation $\displaystyle \frac{dY}{dX}= \frac{4X- Y}{2X+ Y}$.

As to how I went from $\displaystyle \frac{dy}{dx}$ to $\displaystyle \frac{dY}{dx}$, first I replaced y by Y+ 3: $\displaystyle \frac{dY}{dx}= \frac{Y+ 3}{dx}= \frac{dY}{dx}+ 0= \frac{dY}{dx}$. Now to go from $\displaystyle \frac{dY}{dx}$ to $\displaystyle \frac{dY}{dX}$, use the chain rule: $\displaystyle \frac{dY}{dx}= \frac{dY}{dX}\frac{dX}{dx}$ and $\displaystyle \frac{dX}{dx}= 1$ because X= x- 1.

Dividing both numerator and denominator of that fraction by X gives $\displaystyle \frac{dY}{dX}= \frac{4-\frac{Y}{X}}{2+ \frac{Y}{X}}$. Let V= Y/X. Then Y= XV and $\displaystyle \frac{dY}{dX}= V+ X\frac{dV}{dX}= \frac{4- V}{2+ V}$. Then $\displaystyle \frac{2+ V}{4- V}\frac{dV}{dX}= dX$.