How do you solve #5(x-3)+2(3-x)=12#?

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Answer 1 · 2016-02-21T17:56:21

#x=7#

#5(x−3)+2(3−x)=12 #

#5 xx (x) + 5 xx (−3)+2 xx (3) + 2 xx (−x)=12 #

#5x -15 + 6 -2x=12 #

#5x -9 -2x=12 #

#3x -9 =12 #

#3x =12 +9 #

#3x =21#

#x =21/3#

#x=7#

Answer 2 · 2016-02-23T14:22:10

#x=7#

#color(blue)(5(x-3)+2(3-x)=12#

Use distributive property #color(red)(a(b+c)=ab+ac#

#rarr(5x-15)+(6-2x)=12#

Remove the brackets

#rarr5x-15+6-2x=12#

#rarr3x-9=12#

Add #9# both sides

#rarr3x=12+9#

#rarr3x=21#

#rArrcolor(green)(x=21/3=7#