How do you solve #2(y+2)+y=19-(2y+3)#?

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How do you solve #2(y+2)+y=19-(2y+3)#?

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Answer 1 · 2015-03-25T03:32:25

First get rid of the brackets multiplying:
#2y+4+y=19-2y-3#
Isolate #y# on the left and numbers on the right:
#2y+y+2y=-4+19-3#
Add and subtract;
#5y=12#
Divide by #5# both sides to get:
#y=12/5#

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How do you solve #2(y+2)+y=19-(2y+3)#?

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