How do you solve #abs(2t-3) = t# and find any extraneous solutions?
Squaring typically introduces superfluous solutions, but it's worth it because it reduces the complexity of the problem to simple algebra and does away with the perplexing case analysis that's usually involved in an absolute value question.
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To solve the equation abs(2t - 3) = t and find any extraneous solutions, first, consider two cases: when the expression inside the absolute value is positive and when it's negative. Then, solve each case separately and check for extraneous solutions.
Case 1: 2t - 3 ≥ 0 2t ≥ 3 t ≥ 3/2
Case 2: 2t - 3 < 0 2t < 3 t < 3/2
Combine both cases: t ≥ 3/2 or t < 3/2
However, we need to check for extraneous solutions by plugging each solution into the original equation and verifying if it holds true.
For t ≥ 3/2: abs(2(3/2) - 3) = 3/2 abs(3 - 3) = 3/2 abs(0) = 3/2 0 = 3/2 (false)
For t < 3/2: abs(2(1) - 3) = 1 abs(2 - 3) = 1 abs(-1) = 1 1 = 1 (true)
Therefore, the solution to the equation is t < 3/2, and there are no extraneous solutions.
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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.
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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