How do you solve #\sqrt { 2y } - 1= 5#?

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Answer 1 · 2017-05-19T22:30:08

#y=18#

Begin by adding #1# to both sides:

#sqrt(2y)cancel(-1+1)=5+1#

#sqrt(2y)=6#

Apply the square root property:

#sqrt(2y)^2=6^2#

#2y=36#

Divide both sides by #2#

#cancel(2/2)y=36/2#

#y=18#