How do you isolate #r# in #4(x-3)-r+2x=5(3x-7)-9x#?

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Answer 1 · 2016-08-21T14:32:20

#r=23#

#4(x-3)-r+2x=5(3x-7)-9x#

or

#4x-12-r+2x=15x-35-9x#

or

#6x-12=r+6x-35#

or

#r-35=-12#

or

#r=35-12#

or

#r=23#

Answer 2 · 2016-08-21T14:44:07

#r=23#

#4(x-3)-r+2x=5(3x-7)-9x#

#hArr4x-12-r+2x=15x-35-9x#

#hArr6x-12-r=6x-35#

or #cancel(6x)-12-r=cancel(6x)-35#

or #-12-r=-35# or

#35=12+r# or

#r=35-12=23#