Dear community members,
I am out of school for more than 10 years and now have to transpose an equation for my work to $x$. I already tried to find a solution, but failed. I would be very grateful for some explanations.
$$y = \frac{(Bx)}{(K+x)} + Nx$$
Thank you in advance,
Anton
On
The first thing I would do is subtract Nx from both sides: $y- Nx= \frac{Bx}{K+ x}$.
Now get rid of that fraction by multiplying both sides by K+ X: $(y- Nx)(K+ x)= Ky-KNx+ xy- Nx^2= Bx$.
Combine all terms on the left side: $Bx- KNx- xy- Nx^2+ Ky= 0$
Just because I like my leading term to be positive, multiply both sides by -1: $Nx^2+ (KN- B- y)x- Ky- 0$.
Finally, since this involves "$x^2$" use the "Quadratic Formula" to solve for x: $x= \frac{-KN+ B+ Y\pm\sqrt{(KN-B-y)^2+ 4NKy}}{2N}$
$$y=\frac{Bx}{K+x}+Nx$$ $$y(K+x)=Bx+Nx(K+x)$$ $$yK+yx=Bx+NKx+Nx^2$$ $$Nx^2+(NK+B-y)x-Ky=0$$
Now use the quadratic formula:
$$x=\frac{-NK-B+y\pm\sqrt{(NK+B-y)^2+4NKy}}{2N}$$
(there will be up to two solutions for $x$, taking the sign $+$ or $-$ in the numerator).