How do you solve #4(3r - 2) = - 3( r + 7)#?

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Answer 1 · 2017-09-05T00:30:41

#r=-13/15#

Begin by distributing the #4# to #(3r-2)# and the #-3# to #(r+7)#

#[color(red)4*3r]+[color(red)4*-2]=[color(red)(-3)*r]+color(red)[(-3)*7]#

#12r-8=-3r-21#

Add #8# and #3r# to both sides

#12r+color(red)(3r)cancel(-8)cancelcolor(blue)(+8)=cancel(-3r)cancelcolor(red)(+3r)-21color(blue)(+8)#

#15r=-13#

Divide #15# from both sides

#cancel15/cancelcolor(red)15r=-13/color(red)15#

#r=-13/15#