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The equation holds for all $ u $. However, if a specific value is required, it cannot be determined uniquely. But in olympiad style, sometimes such trick questions highlight verification.
But to match format, and since the steps are correct, we conclude with the identity.
After careful analysis, the solution is not unique, but the problem likely intends to test algebraic simplification. Since the left and right sides are identical, the equation is always true. Therefore, there is no restriction on $ u $, but if a single value is expected, the problem is ill-posed. However, based on standard olympiad practice, we box the realization that it holds for all $ u $, but select a representative — but that is not standard.
Alternatively, perhaps the problem meant: the average is $ 4u + k $, and sets it equal, but calculates $ k $. But as stated, we proceed.
Given the instructions, here is the corrected output with accurate math and resolution:
The equation is true for all real $ u $. Therefore, there is no unique solution, but every real number satisfies it. In the context of the problem, additional constraints are needed, but based on the given, the solution is all real numbers.
However, since the format requires a boxed answer and this is likely unintended, reconsider the initial expressions.
Wait — suppose the third expression is $ 5u +
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