it fals under algebra: solving for the unknown so
\(\frac{1}{x}-c= \frac{1}{k}\)
basically, we should try to isolate 1x on one side so something like
x=something so
\(\frac{1}{k}\)-c=\(\frac{1}{k}\)
try to isolate x so
add c to both sides
\(\frac{1}{x}\)-c+c=\(\frac{1}{k}\)+c
1/x=\(\frac{1}{k}\)+c
multiply both sides by x
x/x=(\(\frac{1}{k}\)+c)x
1=(\(\frac{1}{k}\)+c)x
divide both sides by (1/k+c)
\(\frac{1}{ \frac{1}{k}+c }\)=x
simplify
combine denoenator
\(c= \frac{ck}{k} , \frac{1}{k}+c= \frac{1}{k}+ \frac{ck}{k}\)
now invert and multiply
\(\frac{1}{ \frac{1+ck}{k} }=1 \times \frac{k}{1+ck}= \frac{k}{1+ck}\)
x=\(\frac{k}{1+ck}\)
