How do you solve #5 /( y2) = y+2 # and find any extraneous solutions?
First, let us identify any restrictions on the variable.
Therefore, we can find any restrictions by setting the denominator to 0 and solving.
Solving:
Hopefully this helps!
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To solve the equation 5/(y2) = y+2 and find any extraneous solutions, we can follow these steps:

Start by multiplying both sides of the equation by (y2) to eliminate the denominator: 5 = (y+2)(y2)

Expand the right side of the equation: 5 = y^2  4

Rearrange the equation to bring all terms to one side: y^2  4  5 = 0

Simplify the equation: y^2  9 = 0

Factor the equation: (y3)(y+3) = 0

Set each factor equal to zero and solve for y: y3 = 0 > y = 3 y+3 = 0 > y = 3

Check for extraneous solutions by substituting the found values back into the original equation: For y = 3: 5/(32) = 3+2 5/1 = 5 The equation holds true.
For y = 3: 5/(32) = 3+2 5/(5) = 1 The equation does not hold true.

Therefore, the solution to the equation is y = 3, and the extraneous solution is y = 3.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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