How do you solve #3(x-2)+2(x+5)=5(x+1)+1#?

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Answer 1 · 2017-01-09T19:43:40

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There are no values of #x# which are a solution to the problem therefore #x# is the null set or #x = {O/}#

Expand the terms in parenthesis:

#3x - 6 + 2x + 10 = 5x + 5 + 1#

#5x + 4 = 5x + 6#

Subtract #5x# from each side of the equation:

#5x - 5x + 4 = 5x -5x + 6#

#0 + 4 = 0 + 6#

#4 = 6#

Because #4 != 6# there are no values of #x# which are a solution to the problem therefore #x# is the null set or #x = {O/}#