How do you solve #5/(y-2) = y+2#?

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Answer 1 · 2015-06-04T09:15:00

#5/(y-2) = y+2#

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

We know the Difference of Squares Identity which says:
#color(blue) ((a+b)(a-b) = (a^2 - b^2) #

#color(blue)((y+2)(y-2) = y^2 - 2^2 = y^2-4#

the expression now becomes:

#5 = y^2-4#
#9 = y^2#
# y = sqrt9 = color(green)( +3 or -3#

the solutions for the expression are: # color(green)(y = +3 , y = -3#