Х-2у^=1 х-8у=1 поможіть розвязати систему

Решите задачу:

$\left \{ {{x-2y^{2}=1} \atop {x-8y=1}} \right. \\ \left \{ {{1+8y-2y^{2}-1=0} \atop {x=1+8y}} \right. \\ \left \{ {{8y-2y^{2}=0} \atop {x=1+8y}} \right. \\ \\ \left \{ {{2y(4-y)=0} \atop {x=1+8y}} \right. \\$
$\left \{ {{2y=0} \atop { 4-y=0}} \atop {x=1+8y}} \right. \\ \\ y_{1}=0 \\ y_{2}=4 \\ x_{1}=1+8*0 \\ x_{2}=1+8*4 \\ \\ y_{1}=0 \\ y_{2}=4 \\ x_{1}=1+0\\ x_{2}=1+32 \\$

$\\ y_{1}=0 \\ y_{2}=4 \\ x_{1}=1\\ x_{2}=33 \\ \\$