How do you solve #|3x + 2| = 5# and find any extraneous solutions?

Question

Hazel Anglin · Accepted Answer

See a solution process below:

We must solve the term within the absolute value function for both its negative and positive equivalent because the absolute value function takes any term, whether positive or negative, and converts it to its positive form. First Solution #3x + 2 = -5# #3x + 2 - color(red)(2) = -5 - color(red)(2)# #3x + 0 = -7# #3x = -7# #(3x)/color(red)(3) = -7/color(red)(3)# #(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = -7/3# #x = -7/3# Option 2) #3x + 2 = 5# #3x + 2 - color(red)(2) = 5 - color(red)(2)# #3x + 0 = 3# #3x = 3# #(3x)/color(red)(3) = 3/color(red)(3)# #(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 1# #x = 1# The solutions are: #x = -7/3# and #x = 1# Since both solutions address the issue, none of them is superfluous.

Ethan Adkins · Answer

undefined To solve $ |3x + 2| = 5 $, you split it into two cases:

1. $ 3x + 2 = 5 $ 
2. $ 3x + 2 = -5 $

Solve each case separately:

1. $ 3x + 2 = 5 $ 
   Subtract 2 from both sides: $ 3x = 3 $ 
   Divide both sides by 3: $ x = 1 $

2. $ 3x + 2 = -5 $ 
   Subtract 2 from both sides: $ 3x = -7 $ 
   Divide both sides by 3: $ x = -\frac{7}{3} $

Check for extraneous solutions by substituting both solutions into the original equation:

1. For $ x = 1 $: $ |3(1) + 2| = 5 $ 
   $ |3 + 2| = 5 $ 
   $ |5| = 5 $ 
   The equation holds true.

2. For $ x = -\frac{7}{3} $: $ |3(-\frac{7}{3}) + 2| = 5 $ 
   $ |-7 + 2| = 5 $ 
   $ | -5 | = 5 $ 
   The equation holds true.

There are no extraneous solutions.