How do you solve #3=2+2/(z+2)#?

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Answer 1 · 2018-03-05T00:17:06

See a solution process below:

First, subtract #color(red)(2)# from each side of the equation to isolate the #z# term while keeping the equation balanced:

#3 - color(red)(2) = 2 - color(red)(2) + 2/(z + 2)#

#1 = 0 + 2/(z + 2)#

#1 = 2/(z + 2)#

or

#1/1 = 2/(z + 2)#

Because both sides of the equation are pure fractions we can "flip" or invert the fractions without impact the solution:

#1/1 = (z + 2)/2#

#1 = (z + 2)/2#

Next, multiply each side of the equation by #color(red)(2)# to eliminate the fraction while keeping the equation balanced:

#color(red)(2) xx 1 = color(red)(2) xx (z + 2)/2#

#2 = cancel(color(red)(2)) xx (z + 2)/color(red)(cancel(color(black)(2)))#

#2 = z + 2#

Now, subtract #color(red)(2)# from each side of the equation to solve for #z# while keeping the equation balanced:

#2 - color(red)(2) = z + 2 - color(red)(2)#

#0 = z + 0#

#0 = z#

#z = 0#