How do you solve #(2x+1)(1/3)=3# and find any extraneous solutions?

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Answer 1 · 2017-05-02T14:00:42

See the entire solution process below:

First, multiply each side of the equation by #color(red)(3)# to eliminate the fraction while keeping the equation balanced:

#(2x + 1)(1/3) xx color(red)(3) = 3 xx color(red)(3)#

#(2x + 1)(1/color(red)(cancel(color(black)(3)))) xx cancel(color(red)(3)) = 9#

#2x + 1 = 9#

Next, subtract #color(red)(1)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#2x + 1 - color(red)(1) = 9 - color(red)(1)#

#2x + 0 = 8#

#2x = 8#

Now, divide each side of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#(2x)/color(red)(2) = 8/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = 4#

#x = 4#

There are no extraneous solutions.