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

Question

Ethan Adkins · Accepted Answer

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# No superfluous solutions exist.

Julia Adams · Answer

undefined To solve $ (2x + 1)\left(\frac{1}{3}\right) = 3 $, first, multiply both sides of the equation by 3 to eliminate the fraction. This yields $ 2x + 1 = 9 $. Then, subtract 1 from both sides to isolate the variable, giving $ 2x = 8 $. Finally, divide both sides by 2 to solve for $ x $, resulting in $ x = 4 $. To check for extraneous solutions, substitute $ x = 4 $ back into the original equation. If it satisfies the original equation, it is a valid solution; if not, it is extraneous. In this case, $ (2(4) + 1)\left(\frac{1}{3}\right) = 9/3 = 3 $, so $ x = 4 $ is a valid solution, and there are no extraneous solutions.

Julia Adams · Answer

undefined The solution to the equation is x = 4. However, there are no extraneous solutions.