1.2 Solve for \( x \) and \( y \) simultaneously: \( x+2=2 y \) and \( \frac{1}{x}+\frac{1}{y}=1 \)

To solve for \( x \) and \( y \) simultaneously using the equations \( x + 2 = 2y \) and \( \frac{1}{x} + \frac{1}{y} = 1 \), we can start by rearranging the first equation: 1. Rearranging \( x + 2 = 2y \) gives \( x = 2y - 2 \). Now, substitute \( x \) into the second equation: 2. Substitute to get \( \frac{1}{2y - 2} + \frac{1}{y} = 1 \). To find a common denominator, multiply through by \( y(2y - 2) \): 3. \( y + 2y - 2 = y(2y - 2) \) 4. Combine like terms: \( 3y - 2 = 2y^2 - 2y \). Rearranging gives: 5. \( 2y^2 - 5y + 2 = 0 \). Now we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 2, b = -5, c = 2 \): 6. This yields \( y = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} \). Calculating the two possible values for \( y \): 7. \( y = \frac{8}{4} = 2 \) and \( y = \frac{2}{4} = \frac{1}{2} \). Next, use the values of \( y \) to find corresponding values of \( x \): 8. For \( y = 2: x = 2(2) - 2 = 2 \) and for \( y = \frac{1}{2}: x = 2\left(\frac{1}{2}\right) - 2 = -1 \). The solutions are \( (x, y) = (2, 2) \) and \( (x, y) = (-1, \frac{1}{2}) \).