# How do you solve and check for extraneous solutions in #4/v + 1/5 = 1#?

Isolate the term containing the variable; multiply to clear the denominators and divide by the constant to get

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To solve the equation 4/v + 1/5 = 1 and check for extraneous solutions, follow these steps:

- Multiply the entire equation by the least common denominator (LCD) of the fractions, which is 5v. This step eliminates the denominators.

5v * (4/v) + 5v * (1/5) = 5v * 1

- Simplify the equation by distributing and canceling out terms:

20 + v = 5v

- Move all terms involving v to one side of the equation:

v - 5v = -20

-4v = -20

- Divide both sides of the equation by -4 to isolate v:

v = -20 / -4

v = 5

- Substitute the found value of v back into the original equation to check for extraneous solutions:

4/5 + 1/5 = 1

This equation is true, so the solution v = 5 is valid and not extraneous.

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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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