2007-04-27 20:08:42 · 10 answers · asked by Vokalskilz 1 in Science & Mathematics ➔ Mathematics
simple
1/x = 1/y - 1/z
The above equation can be written as
1/z = 1/x - 1/y (sign change in 1/z & 1/x since the r moved frm RHS to LHS & LHS to RHS respectively)
Take LCM & solve the RHS,
1/z = (y - x)/xy
This can be reciprocated and rewritten as
z = xy/(y-x)
Hence the value of z is xy/(y-x)
2007-04-27 22:18:44 · answer #1 · answered by chan_l_u 2 · 0⤊ 0⤋
z = 1/(1/x - 1/y)
2016-05-20 22:04:20 · answer #2 · answered by ? 3 · 0⤊ 0⤋
1/x=1/y -1/z
=>1/z=1/y -1/x
=>1/z=(x-y)/xy
=> z=xy/(x-y)
2007-04-27 21:41:02 · answer #3 · answered by alpha 7 · 0⤊ 0⤋
1/x = 1/y - 1/z
Multiply though by xyz :
yz = xz - xy
Add xy to both sides :
yz + xy = xz
Subtract yz from both sides :
xy = xz - yz
Take out the factor z on the right-hand side :
xy = z(x - y)
Divide both sides by (x - y) :
z = xy / (x - y)
2007-04-27 22:56:29 · answer #4 · answered by falzoon 7 · 0⤊ 0⤋
1/x = 1/y - 1/z
1/z = - (1/x -1/y)
1/z = (x - y) / xy
z = xy / (x - y)
2007-04-27 20:16:09 · answer #5 · answered by Tubby 5 · 0⤊ 0⤋
1 / z = 1 / y - 1 / x
1 / z = (x - y) / xy
z = (xy) / (x - y)
2007-04-27 20:18:10 · answer #6 · answered by Como 7 · 0⤊ 0⤋
1/x - 1/y = 1/z
(y-x)/xy = 1/z
z = xy/(y-x)
Ilusion
2007-04-29 14:05:08 · answer #7 · answered by Ilusion 4 · 0⤊ 0⤋
1/x +1/z=1/y
1/z=1/y-1/x
1 ={1/y-1/x}z
1[y-x]=z
z =y-z
2007-04-27 20:37:45 · answer #8 · answered by Trika mathematica 1 · 0⤊ 1⤋
z = 1/((1/y) - (1/x))
HTH
Doug
2007-04-27 20:11:54 · answer #9 · answered by doug_donaghue 7 · 0⤊ 2⤋
1/z = 1/y-1/x
1/z = [x-y]/xy
xy/z[x-y] = 1
xy/[x-y] = z
2007-04-27 20:19:47 · answer #10 · answered by Kuan T 2 · 1⤊ 0⤋