How do you solve #5 /( y-2) = y+2 # and find any extraneous solutions?

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Answer 1 · 2016-06-12T19:35:05

First, let us identify any restrictions on the variable.

In rational equations, restrictions make the denominator be equal to #0#.

Therefore, we can find any restrictions by setting the denominator to 0 and solving.

#y - 2= 0#

#y = 2#

Thus, #y !=2#.

Solving:

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

#5 = y^2 - 4#

#9 = y^2#

#+-3 = y#

Neither of these contradict the restrictions, therefore our solution set is #{y = +-3}#.

Hopefully this helps!