How do you solve #4(y - 2) = 3(y + 1) + 1 - 3y#?

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Answer 1 · 2016-01-01T17:10:24

#color(blue)(y=3#

#4(y-2) = 3(y+1) +1 -3y#

#4xx(y)+ 4xx(-2) = 3xx(y) +3xx(1) +1 -3y#

#4y -8 = 3y +3 +1 -3y#

#4y -8 = 3y +4 -3y#

#4y -8 -4 = 3y -3y#

#4y -12 = 0#

#4y = 12#

#y = 12/4#

#color(blue)(y=3#