How do you solve #x-2(x+10)=12#?

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Answer 1 · 2018-05-10T21:37:53

#x=-32#

Expand the brackets

#x-2(x + 10)=12 => x-2x-20=12#

#-x-20=12#

Add #x# to both sides

#-20=12+x #

Subtract 12 from both sides

#-32=x