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How do you solve #abs(2z -3) = 4z - 1#?

1 Answer

Jun 24, 2016

#x=3/2#

Explanation:

It is equivalent to:

#(2z-3>=0 and 2z-3=4z-1) or (2z-3<0 and -2z+3=4z-1)#

#(z>=3/2 and 2z=-2) or (z<3/2 and 6z=4)#

#(z>=3/2 and z=-1) or (z<3/2 and z=2/3)#

the first condition is impossible, then since #2/3<3/2#:

the solution is #x=3/2#

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