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

Answer 1

Abigail Allen

#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#

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Answer 2

Isaiah Anthony

To solve the equation 5/(y-2) = y+2, you can start by multiplying both sides of the equation by (y-2) to eliminate the denominator. This gives you 5 = (y+2)(y-2). Expanding the right side of the equation, you get 5 = y^2 - 4. Rearranging the equation, you have y^2 - 4 - 5 = 0, which simplifies to y^2 - 9 = 0. Factoring the quadratic equation, you have (y-3)(y+3) = 0. Setting each factor equal to zero, you get y-3 = 0 or y+3 = 0. Solving for y, you find y = 3 or y = -3. Therefore, the solutions to the equation are y = 3 and y = -3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.