How do you solve #1/(6a) + 1/(3a^2) = 1/(3a)# and find any extraneous solutions?
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To solve the equation 1/(6a) + 1/(3a^2) = 1/(3a) and find any extraneous solutions, we can follow these steps:

Find a common denominator for the fractions on the left side of the equation. In this case, the common denominator is 6a^2.

Rewrite each fraction with the common denominator:
 1/(6a) becomes (a)/(6a^2)
 1/(3a^2) becomes (2)/(6a^2)

Combine the fractions on the left side of the equation: (a)/(6a^2) + (2)/(6a^2) = 1/(3a)

Combine the numerators over the common denominator: (a + 2)/(6a^2) = 1/(3a)

Crossmultiply to eliminate the denominators: 3a * (a + 2) = 6a^2 * 1

Simplify the equation: 3a^2 + 6a = 6a^2

Move all terms to one side of the equation: 6a^2  3a^2  6a = 0

Combine like terms: 3a^2  6a = 0

Factor out common terms: 3a(a  2) = 0

Set each factor equal to zero and solve for 'a': a = 0 or a  2 = 0

Solve for 'a': a = 0 or a = 2

Check for extraneous solutions by substituting each value back into the original equation:
 For a = 0: 1/(60) + 1/(30^2) = 1/(3*0) > Division by zero is undefined, so a = 0 is an extraneous solution.
 For a = 2: 1/(62) + 1/(32^2) = 1/(3*2) > 1/12 + 1/12 = 1/6 > 1/6 = 1/6, which is true.
Therefore, the only valid solution is a = 2, and a = 0 is an extraneous solution.
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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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