How to make z the subject of the formula?

#z=(y(z-y))/x#

2 Answers
Aviv S.
Feb 17, 2018

#z=(-y^2)/(x-y)#

Explanation:

Expand the expressions and then factor out #z#:

#z=(y(z-y))/x#

#z=(yz-y^2)/x#

#zx=yz-y^2#

#zx-yz=-y^2#

#z(x-y)=-y^2#

#z=(-y^2)/(x-y)#

Or, if you would like to rearrange the terms:

#z=(-1* y^2)/(x-y)#

#z=-1* (y^2)/(x-y)#

#z=(y^2)/(-1* (x-y))#

#z=(y^2)/(-x+y)#

#z=y^2/(y-x)#

Gav
Feb 17, 2018

#z#=#-(y^2/{x-y})#

Explanation:

Firstly, multiply #x and z# =#xz#
Then multiply #y(z-y)# which equals to #yz-y^2#
Shift #yz# to the left, #zx-zy#=#-y^2#
Then take out #z#, #z(x-y)#=#-y^2#\
Then divide #-y^2# by #x-y#, #z#=#-(y^2/{x-y})#
So, #z#=#-(y^2/{x-y})#

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