How do you simplify #(4x-12)/(4x)#?

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Answer 1 · 2018-07-18T01:50:09

#color(maroon)(=> 1 - 3 / x#

#(4x - 12) / 4x#

#=> cancel(4x) / cancel(4x) - 12 / (4x)#

#=> 1 - (cancel4 * 3) / (cancel4 * x)#

#color(maroon)(=> 1 - 3 / x#

Answer 2 · 2018-07-18T06:35:44

#1-3/x#

#4x-12\to4(x-3)#

#\therefore(4x-12)/(4x)=(4(x-3))/(4x)=(\cancel{4/4})((x-3)/(x))#
#=(x-3)/x=x/x-3/x=1-3/x#