How do you simplify #(4x-9)/(2x-3)#?

2 Answers
Jun 17, 2018

Answer:

#(4x-9)/(2x-3)=color(blue)(2-3/(2x-3))color(white)"xxxxx"#...but it is not clear that this is a "simplification"

Explanation:

Since this was asked under "Excluded Values for Rational Expression" perhaps it should be noted that
#color(white)("XXX")(4x-9)/(2x-3)# is not defined if #(2x-3)=0 rarr x=3/2#

Jun 17, 2018

Answer:

#2-3/(2x-3)#

Explanation:

Given: #(4x-9)/(2x-3)#

Starting the same process as long division (this one is short)

#color(white)("ddddd.ddddd")4x-9#
#color(magenta)(2)(2x-3)-> ul(4x-6larr" Subtract")#
#color(white)("dddddddddddd")0color(magenta)(-3 larr" Remainder")#

#(4x-9)-:(2x-3) = color(magenta)(2-3/(2x-3))#