How do you solve #|a-4|=3a-6#?

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Answer 1 · 2016-09-29T14:23:14

#a=5/2#

#abs(a-4)=3a-12+6 = 3(a-4)-6# Now, supposing #a ne 4#

#1=3(a-4)/abs(a-4)-6/abs(a-4)# or

#1 = pm3-6/abs(a-4)#

1) #1 = 3-6/abs(a-4)->abs(a-4)=3/2# so
1-a) #-(a-4)=3/2->a=5/2#
1-b) #(a-4)=3/2->a=11/2#

2) #1=-3-6/abs(a-4)# is not feasible

Analysing the outcomes the only feasible is

#a=5/2#