How do you solve the equation #abs(2x-1)=x+1#?

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Answer 1 · 2017-08-15T11:04:54

See below.

#abs(2x-1)=x-1/2+3/2 = 1/2(2x-1)+3/2#

now assuming #2x-1 ne 0# we have

#abs(2x-1)/(2x-1) = 1/2+3/2 1/(2x-1)# so

#2abs(2x-1)/(2x-1) = 1+3/(2x-1)#

but #abs(2x-1)/(2x-1) = pm 1# so we have two options

#{(-2=1+3/(2x-1)->-3=3/(2x-1)->2x-1=-1 -> x = 0),(2=1+3/(2x-1)->1=3/(2x-1)->2x-1=3->x=2):}#

so the solutions are

#x = {0, 2}#