Помогите пожалуйста решить 3 и 4 задание

Длина - x , ширина y
x*y = 300
(x + 5)(y - 5) = 250
xy - 5x + 5y - 25 = 250
xy - 5(x - y) - 25 = 250
300 - 5(x - y) - 25 = 250
- 5(x - y) = - 25
x - y = 5 $\left \{ {{x-y=5} \atop {xy=300}} \right.\\\\ \left \{ {{x=5+y} \atop {(5+y)y=300}}\\\\ \left \{ {{x=5+y} \atop { y^{2}+5y-300=0 }} \right. \right.\\\\ \left \{ {{y _{1} =15} \atop {x _{1} =5+15}} \right.\\\\ \left \{ {{y _{1} =15} \atop {x _{1} =20}} \right.\\\\ \left \{ {{y _{2} =-20} \atop {x _{2}=5-20 }} \right\\\\ \left \{ {{y _{2} =-20} \atop {x _{2}=-15 }} \right. .$
Ответ: (20 ;15) , (- 15 ; - 20)

$\left \{ {{(x + 3y) ^{2}=16 } \atop {x=3y-2}} \right. \\\\ \left \{ {{x+3y =4} \atop {x=3y-2}} \right.\\\\ \left \{ {{3y-2+3y=4} \atop {x=3y-2}} \right.\\\\ \left \{ {{6y=6} \atop {x=3y-2}} \right.\\\\ \left \{ {{y=1} \atop {x=1}} \right. \\\\ \left \{ {{x+3y=-4} \atop {x=3y-2}} \right.\\\\ \left \{ {{3y-2+3y=-4} \atop {x=3y-2}} \right.\\\\ \left \{ {{6y=-2} \atop {x=3y-2}} \right.\\\\ \left \{ {{y=- \frac{1}{3} } \atop {x=-3}} \right.$
Ответ. (1 , 1),(- 3 ; - 1/3)