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

2 Answers
Apr 4, 2018

Answer:

#y = +-3#

Explanation:

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

First of all, we multiply both sides by #y-2#:
#5 = (y+2)(y-2)#

#5 = y^2 - 4#

#9 = y^2#

#y = +-3#

Now we have to check both of our solutions back into the original equation to make sure both are really solutions:
#5/(-3-2)=-3+2#

#5/-5 = -1#

#-1 = -1#

Now we know that #-3# works. Now let's check #3#:
#5/(3-2) = 3+2#

#5/1 = 5#

#5 = 5#

Good. Both solutions work. So the answer is #y = +-3#.

Hope this helps!

Apr 4, 2018

Answer:

#y = +-3#

Explanation:

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

Multiply #(y - 2)# on both the sides

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

#5 = (y + 2) × (y - 2)#

#5 = y^2 - 2^2 color(white)(..)[∵ (a + b)(a - b) = a^2 - b^2)#

#5 = y^2 - 4#

Add #4# on both the sides

#5 + 4 = y^2 - 4 + 4#

#9 = y^2#

Square root on both the sides

#sqrt(9) =sqrt(y^(2))#

#color(blue)(y = +-3)#