How do you solve #abs(4-3x)= 4x + 6# and state the extraneous solutions?

Question

Eli Assouline · Accepted Answer

#abs(4-3x) = 4x+6# Consider the two possibilities for #(4-3x)# Possibility 1: #(4-3x)<0# #rarr x>4/3# #abs(4-3x) = 4x +6# becomes #3x-4 = 4x+6# #x=-10# but #x>4/3# for this condition to apply so this solution is extraneous. Possibility 2: #(4-3x)>=0# #rarr x<= 4/3# #abs(4-3x)= 4x+6# becomes #4-3x = 4x+6# #x = 2/7# Since #x=2/7# satisfies the condition #x<= 4/3# this is a valid solution

Kennedy Adams · Answer

undefined To solve the equation $ |4 - 3x| = 4x + 6 $ and identify any extraneous solutions, follow these steps:

1. Split the equation into two cases: $ 4 - 3x = 4x + 6 $ and $ 4 - 3x = -(4x + 6) $.
2. Solve each case separately.
3. Check each solution to see if it satisfies the original equation.

Case 1: $ 4 - 3x = 4x + 6 $
$$ -3x - 4x = 6 - 4 $$
$$ -7x = 2 $$
$$ x = -\frac{2}{7} $$

Case 2: $ 4 - 3x = -(4x + 6) $
$$ -3x - 4x = -6 - 4 $$
$$ -7x = -10 $$
$$ x = \frac{10}{7} $$

Check each solution:
For $ x = -\frac{2}{7} $:
$$ |4 - 3\left(-\frac{2}{7}\right)| = 4\left(-\frac{2}{7}\right) + 6 $$
$$ |4 + \frac{6}{7}| = -\frac{8}{7} + 6 $$
$$ \frac{34}{7} = \frac{34}{7} $$ (True)

For $ x = \frac{10}{7} $:
$$ |4 - 3\left(\frac{10}{7}\right)| = 4\left(\frac{10}{7}\right) + 6 $$
$$ |4 - \frac{30}{7}| = \frac{40}{7} + 6 $$
$$ \frac{2}{7} = \frac{82}{7} $$ (False)

So, the solution to the equation is $ x = -\frac{2}{7} $, and the extraneous solution is $ x = \frac{10}{7} $.