First we need to add and subtract the necessary terms from each side of the equation to isolate the #y# terms on one side of the equation and the constants on the other side of the equation while keeping the equation balanced:

#5y - 2 - color(red)(2y) + color(blue)(2) = 2y + 19 - color(red)(2y) + color(blue)(2)#

#5y - color(red)(2y) - 2 + color(blue)(2) = 2y - color(red)(2y) + 19 + color(blue)(2)#

#(5 - 2)y - 0 = 0 + 21#

#3y = 21#

We can now divide each side of the equation by #color(red)(3)# to solve for #y# while keeping the equation balanced:

#(3y)/color(red)(3) = 21/color(red)(3)#

#(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = 7#

#y = 7#