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Solve for yy.\newline1+2y=y\sqrt{1 + 2y} = \sqrt{y}\newliney=y = _____

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Q. Solve for yy.\newline1+2y=y\sqrt{1 + 2y} = \sqrt{y}\newliney=y = _____
  1. Equation Simplification: We have the equation:\newline1+2y=y\sqrt{1 + 2y} = \sqrt{y}\newlineTo eliminate the square roots, we square both sides of the equation:\newline(\sqrt{1 + 2y})^2 = (\sqrt{y})^2\(\newline\$1 + 2y = y\)
  2. Isolating y: Now we solve for y:\(\newline\)\(1 + 2y = y\)\(\newline\)Subtract \(y\) from both sides to isolate terms with \(y\) on one side:\(\newline\)\(2y - y = -1\)\(\newline\)\(y = -1\)

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