How do you solve #1/x+2/(3x)=1/3# and find any extraneous solutions?
Multiply by three now.
It is not unnecessary since the solution functions.
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To solve the equation 1/x + 2/(3x) = 1/3 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 3x.

Multiply each term by the common denominator to eliminate the fractions. This gives us: 3x * (1/x) + 3x * (2/(3x)) = 3x * (1/3).

Simplify the equation by canceling out common factors. This results in: 3 + 2 = x.

Combine like terms on the left side of the equation: 5 = x.

Therefore, the solution to the equation is x = 5.

To check for extraneous solutions, substitute the found value of x back into the original equation: 1/5 + 2/(3*5) = 1/3.

Simplify the equation: 1/5 + 2/15 = 1/3.

Find a common denominator for the fractions on the left side: 3/15 + 2/15 = 1/3.

Combine the fractions: 5/15 = 1/3.

Simplify the left side: 1/3 = 1/3.

Since the equation is true, the solution x = 5 is valid and there are no extraneous solutions.
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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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