How do you solve #5( 3y - 2) = 20#?

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Answer 1 · 2016-10-06T21:11:13

#y = 2#

First we expand the brackets

#5 (3y-2) = 20 to#

#15y - 10 = 20#

Then add #10# to both sides

#15y = 30#

Finally we divide by #15#

#y = 2#