1.1.4 \( \quad 2 \sqrt{x+2}=x-1 \) Solve for \( x \) and \( y \) simultaneously:

To solve the equation \( 2 \sqrt{x+2} = x - 1 \), first isolate the square root term: 1. Square both sides to eliminate the square root: \( (2 \sqrt{x+2})^2 = (x - 1)^2 \) leads to \( 4(x + 2) = x^2 - 2x + 1 \). 2. Expanding gives: \( 4x + 8 = x^2 - 2x + 1 \). 3. Rearranging terms results in: \( x^2 - 6x - 7 = 0 \). 4. Factoring yields \( (x - 7)(x + 1) = 0 \), giving potential solutions \( x = 7 \) and \( x = -1 \). 5. Substitute these back into \( 2 \sqrt{x + 2} = x - 1 \) to check for validity. Only \( x = 7 \) satisfies the original equation. Now for \( y \), it seems there's a lack of context regarding its definition or relationship with \( x \). Assuming \( y = x \) as a function of the solution, we'd have \( y = 7 \). Therefore, the solution set \( (x, y) = (7, 7) \). To sum up, it's always crucial to check solutions after squaring equations, as extraneous solutions may arise!