How do you solve for #z# in #12- y + z = A#?

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Answer 1 · 2017-07-16T21:08:33

See a solution process below:

First, subtract #color(red)(12)# from each side of the equation to begin to isolate the #z# term while keeping the equation balanced:

#12 - color(red)(12) - y + z = A - color(red)(12)#

#0 - y + z = A - 12#

#-y + z = A - 12#

Next, add #color(red)(y)# to each side of the equation to solve for #z# while keeping the equation balanced:

#color(red)(y) - y + z = color(red)(y) + A - 12#

#0 + z = y + A - 12#

#z = y + A - 12#