How do you solve #3z + 20= 41#?

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Answer 1 · 2017-12-19T12:56:36

See a solution process below:

First, subtract #color(red)(20)# from each side of the equation to isolate the #z# term while keeping the equation balanced:

#3z + 20 - color(red)(20) = 41 - color(red)(20)#

#3z + 0 = 21#

#3z = 21#

Now, divide each side of the equation by #color(red)(3)# to solve for #z# while keeping the equation balanced:

#(3z)/color(red)(3) = 21/color(red)(3)#

#(color(red)(cancel(color(black)(3)))z)/cancel(color(red)(3)) = 7#

#z = 7#