algebraic fractions,formulae

1. Make k be the subject of the formulae 5/h-10=1/5-k

A. (5/h)-10 = 1/(5-k) [k]
5-10h = h/(5-k)
(5-10h)(5-k) = h
25-5k-50h+10hk = h
5k(2h-1) = 51h-25
k = (51h-25)/[5(2h-1)]
or
(5/h)-10 = (1/5)-k
k = (1/5)-(5/h)+10
or
5/(h-10) = 1/(5-k)
5(5-k) = 1*(h-10) ----------------times (5-k)(h-10) each sides
25-5k = h-10
5k = 25-h+10
k = (35-h)/5
or
5/(h-10) = (1/5)-k
k = (1/5)-[5/ (h-10)]

2. Make t the subject of the formula rt-1/r=1-2t/s

A. (rt-1)/r = (1-2t)/s [t]
s(rt-1) = r(1-2t)
srt-s = r-2rt
srt+2rt = r+s
tr(s+2) = r+s
t = (r+s)/[r(s+2)]
or
(rt-1)/r = 1-(2t/s)
[s(rt-1)]/r = s-2t
srt-s = rs-2rt
rt(s+2) = s(r+1)
t = s(r+1)/[r(s+2)]
or
rt-(1/r) = (1-2t)/s
srt-(s/r) = 1-2t
2t = 1-srt+(s/r)
t = (1-srt)/2+(s/2r)
or
rt-(1/r) = 1-(2t/s)
tr^2-1 = r-(2rt/s)
str^2-s = rs-2rt
tr(sr+2) = s(r+1)
t = s(r+1)/[r(sr+2)]

3. Simplify w/w-8+3/8(8-w)

A. w/(w-8)+3/[8(8-w)]
= w/(w-8)-3/[8(w-8)]
= (8w-3)/[8(w-8)]

no one answer you because your questions is not clear ,

For example , 5/h-10=1/5-k

is (5/h) - 10 = (1/5) - k

or

5/(h-10) = 1/(5-k)

??

rt-1/r=1-2t/s

is (rt) - (1/r) = 1 - (2t/s) or (rt-1)/r = (1-2t)/s ???

which one is the subject?