Question

Допоможіть розв'язати.
Даю 44б.​

Answer 1

$\displaystyle\bf\\-\left \{ {{2y^{2} +xy=36} \atop {y^{2} +xy=20}} \right. \\----------\\y^{2} =16\\\\y_{1,2} =\pm \ 4\\\\1)\\\\y_{1} =-4\\\\y^{2} +xy=20\\\\(-4)^{2} +x\cdot(-4)=20\\\\16-4x=20\\\\-4x=4\\\\x_{1} =-1\\\\\\2)\\\\y_{2} =4\\\\y^{2} +xy=20\\\\4^{2} +x\cdot 4=20\\\\16+4x=20\\\\4x=4\\\\x_{2} =1\\\\\\Otvet: \ (-1 \ ; \ -4) \ , \ (1 \ ; \ 4)$