How do you solve #8z=4(2z+1)#?

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Answer 1 · 2017-02-05T11:50:29

See the entire solution process below:

First, expand the terms within parenthesis on the right hand side of the equation:

#8z = (4 xx 2z) + (4 xx 1)#

#8z = 8z + 4#

Now, subtract #8z# from each side of the equation:

#-8z + 8z = -8z + 8z + 4#

#0 = 0 + 4#

#0 = 4#

However, we know #0 != 4#, therefore there is no solution to this equation, or #x# equals the null set, or #x = {O/}#