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How do you solve #abs(5-4w )= 3w# and find any extraneous solutions?

Answer 1

Avery Alexander

#w = 5/7# and #w = 5#

The equation #abs(5-4w )= 3w# is equivalent to the couple of independent equations

#{ (5-4w= 3w), (5-4w= -3w) :}#

Solving we have #w = 5/7# and #w = 5#

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Answer 2

Abigail Allen

To solve (|5-4w| = 3w), consider both the positive and negative cases of the absolute value expression.

Case 1 (Positive case): [5 - 4w = 3w] Combine like terms: [5 = 7w] [w = \frac{5}{7}]

Case 2 (Negative case): [5 - 4w = -3w] Combine like terms: [5 = w] [w = 5]

Check for extraneous solutions:

For (w = \frac{5}{7}): [|5 - 4(\frac{5}{7})| = 3(\frac{5}{7})] [|5 - \frac{20}{7}| = \frac{15}{7}] [|\frac{35}{7} - \frac{20}{7}| = \frac{15}{7}] [|\frac{15}{7}| = \frac{15}{7}] The equation holds, so (w = \frac{5}{7}) is a valid solution.

For (w = 5): [|5 - 4(5)| = 3(5)] [|5 - 20| = 15] [|-15| = 15] The equation holds, so (w = 5) is also a valid solution.

Thus, both (w = \frac{5}{7}) and (w = 5) are solutions, and there are no extraneous solutions.

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Answer from HIX Tutor

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.