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

Answer 1

#y=+-3# is The Soln.

#5/(y-2)=y+2. rArr 5=(y-2)(y+2) rArr 5= y^2-4 rArr 5+4 =y^2 rArr y^2=9 rArr y=+-3#
These roots satisfy the given eqn. Hence, the roos of the eqn. are #+-3#
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Answer 2

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 this 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 5/(y-2) = y+2 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.

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.

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.

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.

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