Ребят помогите пожалуйста. прошу подробное решение

$\left \{ {{4x^{2}+3xy-y^{2}=0} \atop {x^{2}-2xy+y^{2}=16}} \right.$

$\left \{ {{4x^{2}+3xy-y^{2}=0} \atop {(x-y)^{2}=16}} \right.$

$\left \{ {{4x^{2}+3xy-y^{2}=0} \atop {x-y=4, x-y=-4}} \right.$

$\left \{ {{4x^{2}+3xy-y^{2}=0} \atop {x=4-y, x=y-4}} \right.$

при х=4-y
$\left \{ {{4*(4-y)^{2}+3y*(4-y)-y^{2}=0} \atop {x=4-y}} \right.$

$\left \{ {{4*(16-8y+y^{2})+12y-3y^{2}-y^{2}=0} \atop {x=4-y}} \right.$

$\left \{ {{64-32y+4y^{2}+12y-4y^{2}=0} \atop {x=4-y}} \right.$

$\left \{ {{64-20y=0} \atop {x=4-y}} \right.$

$\left \{ {{y=\frac{16}{5}} \atop {x=4-\frac{16}{5}=\frac{4}{5}}} \right.$

при х=y-4
$\left \{ {{4*(y-4)^{2}+3y*(y-4)-y^{2}=0} \atop {x=y-4}} \right.$

$\left \{ {{4y^{2}-32y+64+3y^{2}-12y-y^{2}=0} \atop {x=y-4}} \right.$

$\left \{ {{6y^{2}-44y+64=0} \atop {x=y-4}} \right.$

0} \atop {x=y-4}} \right." alt=" \left \{ {{3y^{2}-22y+32=0, D=100>0} \atop {x=y-4}} \right." align="absmiddle" class="latex-formula">

$\left \{ {{y_{1}= \frac{22-10}{6}=2} \atop {x=2-4=-2}} \right.$

$\left \{ {{y_{2}=\frac{16}{3}} \atop {x=\frac{16}{3}-4=\frac{4}{3}}} \right.$

Ответ: (4/5; 16/5); (-2; 2); (4/3; 16/3)