(a) Prove that for every real number x there is a unique real number y such that y ? 0 and y^2 - x^2 - 1 = 0. (b) Prove that there exists a unique real number y such that for every real number x, 2yx + 6 = 4x + 3y.
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Since the square root of a positive number is always positive, we have two possible solutions for \(y\): \(y = \sqrt{x^2 + 1}\) or \(y = -\sqrt{x^2 + 1}\). **Step 2:** Now, let's consider the equation \(2yx + 6 = 4 + 3y\). **Step 3:** Rearrange the equation to Show more…
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