One process is to first subtract #color(red)(9/y)# from each side of the equation to isolate all the #y# terms on on side of the equation while keeping the equation balanced:
#-color(red)(9/y) + 9/y + 2/3 = -color(red)(9/y) + 3/(3y)#
#0 + 2/3 = -9/y + color(blue)(cancel(color(black)(3)))/(color(blue)(cancel(color(black)(3)))y)#
#2/3 = -9/y + 1/y#
#2/3 = (-9 + 1)/y#
#2/3 = (-8)/y#
Next, multiply each side of the equation by #color(red)(y)# to get the #y# term in the numerator while keeping the equation balanced:
#color(red)(y) xx 2/3 = color(red)(y) xx (-8)/y#
#2/3y = cancel(color(red)(y)) xx (-8)/color(red)(cancel(color(black)(y)))#
#2/3y = -8#
No, multiply each side of the equation by #color(red)(3)/color(blue)(2)# to solve for #y# while keeping the equation balanced:
#color(red)(3)/color(blue)(2) xx 2/3y = color(red)(3)/color(blue)(2) xx -8#
#cancel(color(red)(3))/cancel(color(blue)(2)) xx color(blue)(cancel(color(black)(2)))/color(red)(cancel(color(black)(3)))y = -24/2#
#y = -12#
