How do you solve for #z# in #xz+y=1+z#?

1 Answer
Mar 17, 2018

Answer:

See a solution process below:

Explanation:

First, subtract #color(red)(y)# and #color(blue)(z)# from each side of the equation to isolate the #z# terms while keeping the equation balanced:

#xz - color(blue)(z) + y - color(red)(y) = 1 - color(red)(y) + z - color(blue)(z)#

#xz - z + 0 = 1 - y + 0#

#xz - z = 1 - y#

Next, factor a #z# out of each term on the left giving:

#z(x - 1) = 1 - y#

Now, divide each side of the equation by #color(red)(x - 1)# to solve for #z# while keeping the equation balanced:

#(z(x - 1))/color(red)(x - 1) = (1 - y)/color(red)(x - 1)#

#(zcolor(red)(cancel(color(black)((x - 1)))))/cancel(color(red)(x - 1)) = (1 - y)/(x - 1)#

#z = (1 - y)/(x - 1)#