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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

Jessie
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ASe1988
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  • Welcome to MSE! What attempts have you done so far? Where are you stuck? Please use mathjax to show your work and review the meta read for tips on providing context in your question to better assist the community. – Jessie Mar 25 '21 at 12:11

2 Answers2

0

$$y=\frac{Bx}{K+x}+Nx$$ $$y(K+x)=Bx+Nx(K+x)$$ $$yK+yx=Bx+NKx+Nx^2$$ $$Nx^2+(NK+B-y)x-Ky=0$$

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).

0

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: $x= \frac{-KN+ B+ Y\pm\sqrt{(KN-B-y)^2+ 4NKy}}{2N}$

user247327
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