How do you solve #abs(x + 1)=abs(-4 x + 3)#?

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Answer 1 · 2018-05-25T11:18:16

The solutions are #{(x=4/3),(x=2/5):}#

The equation is

#|x+1|=|-4x+3|#

Removing one absolute value at a time :

#|x+1|=-4x+3# and #|x+1|=4x-3#

The first equation gives

#|x+1|=-4x+3#

#<=>#, #{(x+1=-4x+3),(-x-1=-4x+3):}#

#<=>#, #{(5x=2),(3x=4):}#

#<=>#, #{(x=2/5),(x=4/3):}#

The second equation gives

#|x+1|=4x-3#

#<=>#, #{(x+1=4x-3),(-x-1=4x-3):}#

#<=>#, #{(3x=4),(5x=2):}#

#<=>#, #{(x=4/3),(x=2/5):}#

graph{(y-|x+1|)(y-|-4x+3|)=0 [-6.38, 7.67, -1.74, 5.283]}