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$36) What is the solution set for $ \sqrt{x+11}+1=x $ ? 37) Solve the following system of equations algebraically and chec $ x^{2}+y^{2}=100 $ $ y=x-2 $$

36) What is the solution set for $ \sqrt{x+11}+1=x $ ? 37) Solve the following system of equations algebraically and chec $ x^{2}+y^{2}=100 $ $ y=x-2 $

Ask by Phillips Valdez.

Jan 13,2025 15:29

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**Problem 36:** The solution to the equation $ \sqrt{x + 11} + 1 = x $ is $ x = 5 $. **Problem 37:** The solutions to the system of equations $ x^{2} + y^{2} = 100 $ and $ y = x - 2 $ are the points $ (8, 6) $ and $ (-6, -8) $.

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To solve $ \sqrt{x+11}+1=x $, first isolate the square root: $ \sqrt{x+11} = x - 1 $. Then, square both sides to eliminate the square root, giving $ x + 11 = (x - 1)^2 $. Expanding and simplifying leads to the equation $ x^2 - 3x - 10 = 0 $. Factoring yields $ (x - 5)(x + 2) = 0 $, so $ x = 5 $ or $ x = -2 $. Checking both in the original equation confirms that $ x = 5 $ is valid, but $ x = -2 $ is not. Thus, the solution set is $ \{5\} $. For the second problem, substitute $ y $ in the first equation: $ x^2 + (x - 2)^2 = 100 $. This expands to $ x^2 + x^2 - 4x + 4 = 100 $, simplifying to $ 2x^2 - 4x - 96 = 0 $. Dividing through by 2 yields $ x^2 - 2x - 48 = 0 $, which factors to $ (x - 8)(x + 6) = 0 $. Thus, $ x = 8 $ or $ x = -6 $. Substituting back gives $ y = 6 $ or $ y = -8 $, resulting in the solution pairs $ (8, 6) $ and $ (-6, -8) $.