Решите систему уравнений x-y=4, x^2 + y^2=10.

Решите задачу:

$\left \{ {{x-y=4} \atop {x^{2} -y^{2} =10}} \right. \left \{ {{x=4+y} \atop {x^{2} -y^{2}=10 }} \right. =(4+y)^{2}-y^{2} =10\\16+8y+y^{2}-y^{2} =10 \\16+8y=10\\8y=10-16\\8y=-6:8\\y=-\frac{3}{4}\\\\x=4-\frac{3}{4} \\x=\frac{13}{4} \\(x,y)=(\frac{13}{4},-\frac{3}{4} ) \\\\\left \{ {{\frac{13}{4}-(-\frac{3}{4})=4 } \atop {(\frac{(13}{4} )^{2}-(-\frac{3}{4^{} })^{2} =10 }} \right. \\\\\frac{4=4}{10=10} \\\\(x,y)=(\frac{13}{4}, -\frac{3}{4} )$