[tex]\left \{ {{sin\alpha+cos\alpha =\frac{1}{5} } \atop {sin\alpha -cos\alpha =x}} \right.[/tex]

складываем и вычитаем:

[tex]\left \{ {{2sin\alpha=\frac{1}{5}+x } \atop {2cos\alpha=\frac{1}{5}-x }} \right. \\\\\left \{ {{sin\alpha=\frac{\frac{1}{5}+x}{2} } \atop {cos\alpha=\frac{\frac{1}{5}-x}{2} }} \right.[/tex]

Так как [tex]sin^2x+cos^2x=1[/tex] ,  то

[tex](\frac{\frac{1}{5}+x}{2})^2 +(\frac{\frac{1}{5}-x}{2})^2 =1[/tex]

[tex]\frac{\frac{1}{25}+2\cdot \frac{1}{5}x+x^2 }{4} +\frac{\frac{1}{25}+2\cdot \frac{1}{5}x+x^2 }{4} =1\\\\x^2=\frac{49}{25} \\\\x=\pm\frac{7}{5}[/tex]

О т в е т. [tex]sin\alpha -cos\alpha =\pm\frac{7}{5}[/tex]