Question

(x+y√2)(3-√2)= -2
| x-y = ?

Answer 1

$(x+y \sqrt{2})(3 - \sqrt{2}) = -2 \\ \\ x + y \sqrt{2} = - \frac{2}{(3 - \sqrt{2} )} \\ \\ x + y \sqrt{2} = - \frac{2(3 + \sqrt{2} )}{(3 - \sqrt{2} )(3 + \sqrt{2} )} \\ \\ x + y \sqrt{2} = - \frac{6 + 2 \sqrt{2} }{5} \\ \\ x + y \sqrt{2} = - \frac{6}{5} - \frac{2}{5} \sqrt{2} \\ \\ x = - \frac{6}{5} \\ \\ y = - \frac{2}{5} \\ \\ x - y = - \frac{4}{5} \\ \\ |x - y| = \frac{4}{5}$

Answer 2

[tex](x+y \sqrt{2} )(3- \sqrt{2} )=-2 or,x+y \sqrt{2} = \frac{-2}{3- \sqrt{2} } or,x=\frac{-2}{3- \sqrt{2} } -y \sqrt{2} [/tex]