# How do you solve #3/(x+6)=5/(x-2)# and find any extraneous solutions?

This value is the solution if you substitute it into the equation and both sides come out equal.

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To solve the equation 3/(x+6)=5/(x-2) and find any extraneous solutions, we can start by cross-multiplying to eliminate the fractions. This gives us 3(x-2) = 5(x+6). Expanding and simplifying, we get 3x - 6 = 5x + 30. Rearranging the equation, we have 2x = -36. Dividing both sides by 2, we find x = -18.

To check for extraneous solutions, we substitute x = -18 back into the original equation. We have 3/(-18+6) = 5/(-18-2), which simplifies to 3/(-12) = 5/(-20). Evaluating further, we get -1/4 = -1/4. Since both sides are equal, there are no extraneous solutions in this case. Therefore, the solution to the equation is x = -18.

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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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