How do you solve #2+ \sqrt { 4y } - 3= 11#?

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Answer 1 · 2016-11-11T18:17:44

#y=36#

#2+sqrt(4y) - 3=11#

Simplify.

#sqrt(4y)-1=11#

Add #1# to each side.

#sqrt(4y)=12#

Square both sides.

#4y=144#

Divide both sides by #4#.

#y=36#