How can I find the R values

Question

I have the following equations :

$$\begin{cases}K = \frac{B – 3}{20}\\ K = (20S+3)R+S\\ K = 20S^2 + (20N+7)S + N\\ N= R - S \end{cases}$$ - And I have the $B$ values, e.g : 834343, 3253538, 10^87653

How can I find the $R$ values?

or maybe there is no solution for these equations?

Answer 1

Hint

Hoping that I did not make mistakes, you can immediately eliminate $K$ from the first equation, $N$ from the fourth equation. Replace in the second and third equations; substract one from the other. You will end with a quadratic equation in $R$.

I am sure that you can take from here.

Answer 2

$$K = \frac{B – 3}{20}\tag{1}$$ $$ K = (20S+3)R+S\tag{2}$$ $$ K = 20S^2 + (20N+7)S + N\tag{3}$$ $$ N= R - S \tag{4}$$ From (4), $R=S+N$.

Put in (2): $$K=(20S+3)(N+S)+S=20NS+20S^2+3N+4S\tag{5}$$ Also from (3): $$K=20S^2+20NS+7S+N\tag{6}$$ Subtract (5) from (6): $$K-K=(20S^2+20NS+7S+N)-(20NS+20S^2+3N+4S)\\ 2N=3S$$ Put $N=\frac32S$ in (6) and use (1): $$\frac{B-3}{20}=20S^2+20\left(\frac32S\right)S+7S+\frac32S$$ $$50S^2+\frac{17}2S-\frac{B-3}{20}=0$$ Solve it to get $$S:=f(B)$$ Then $$R=\frac52f(B)$$ Spoiler:Form Answer

$$R = \frac1{80} (\pm\sqrt{40 B+169}-17)$$