How do you solve # abs(2x - 6) = abs(x +7) - 3 #?

Question

Eli Assouline · Accepted Answer

The solutions are #S={2/3,10}#

The equation is

#|2x-6|=|x+7|-3#

#=>#, #|2x-6|-|x+7|+3=0#

The points to be considered are when

#{(2x-6=0),(x+7=0):}#

#=>#, #{(x=3),(x=-7):}#

There are #3# cases to consider

In the interval #(-oo, -7)#

#2x-6-(-x-7)+3=0#

#=>#, #2x-6+x+7+3=0#

#=>#, #3x+4=0#

#=>#, #x=-4/3#

This solution is not possible since #x=-4/3# #!in# to #(-oo,-7)#

In the interval #(-oo, -7)#

#-2x+6-(x+7)+3=0#

#=>#, #-2x+6-x-7+3=0#

#=>#, #-3x+2=0#

#=>#, #x=2/3#

This solution is possible since #x=2/3# #in# to #(-7,3)#

In the interval #(3,+oo)#

#2x-6-(x+7)+3=0#

#=>#, #2x-6-x-7+3=0#

#=>#, #x-10=0#

#=>#, #x=10#

This solution is possible since #x=10# #in# to #(3,+oo)#

The solutions are #S={2/3,10}#

graph{|2x-6|-|x+7|+3 [-18.02, 18.03, -9.01, 9.01]}

Eva Abramson · Answer

undefined To solve the equation $ |2x - 6| = |x + 7| - 3 $, follow these steps:

1. Break down the equation based on the absolute value:
   - For $ |2x - 6| = |x + 7| - 3 $, we have two cases:
     - Case 1: $ 2x - 6 = x + 7 - 3 $ when $ 2x - 6 \geq 0 $ and $ x + 7 - 3 \geq 0 $
     - Case 2: $ 2x - 6 = -(x + 7) - 3 $ when $ 2x - 6 \geq 0 $ and $ -(x + 7) - 3 \geq 0 $

2. Solve each case separately:
   - Case 1:
     $ 2x - 6 = x + 4 $
     $ x = 10 $

- Case 2:
     $ 2x - 6 = -x - 10 $
     $ 3x = -4 $
     $ x = -\frac{4}{3} $

3. Check the solutions to ensure they satisfy the original equation:
   - For $ x = 10 $, $ |2(10) - 6| = |10 + 7| - 3 $ is true.
   - For $ x = -\frac{4}{3} $, $ |2\left(-\frac{4}{3}\right) - 6| = |-\frac{4}{3} + 7| - 3 $ is true.

Therefore, the solutions are $ x = 10 $ and $ x = -\frac{4}{3} $.