[tex]\begin{cases} x+y=4 \\ x^{2} + y^{2}=10 \end{cases} \\ \\\begin{cases} y=4-x \\x^{2} + y^{2}=10 \end{cases} \\\\substitution : \\x^2+(4-x)^2=10\\x^2+16-8x+x^2-10=0\\2x^2 -8x +6=0[/tex]
[tex]x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{8-\sqrt{ (-8)^2-4\cdot 2\cdot 6 }}{2\cdot 2}=\frac{8-\sqrt{64 -48 }}{4}=\frac{8-\sqrt{16 }}{4}= \\=\frac{8-4}{4}=\frac{4}{4}=1\\\\x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{8+\sqrt{ (-8)^2-4\cdot 2\cdot 6 }}{2\cdot 2}= \frac{8+4}{4}=\frac{12}{4}=3[/tex]
[tex]\begin{cases} y=4-x \\ x=1 \end{cases} \ \ or \ \ \begin{cases} y=4-x \\ x=3 \end{cases}\\\\\begin{cases} y=4-1 \\ x=1 \end{cases} \ \ or \ \ \begin{cases} y=4-3 \\ x=3 \end{cases}\\\\\begin{cases} y=3 \\ x=1 \end{cases} \ \ or \ \ \begin{cases} y=1 \\ x=3 \end{cases}[/tex]
[tex]x=1 \ and \ y= 3 \\\\x^4+y^4 = 1^4+3^4=1+ 81=82\\\\ x=3 \ and \ y= 1 \\\\x^4+y^4 = 3^4+1^4= 81+1=82[/tex]