Question

1) x2-y=14
{
3x+y=4

2)2x+5xy=14
{
y-5xy=-9

Answer 1

${x}^{2} - y = 14 \\ 3x + y = 4 \\ \\ {x}^{2} + 3x = 18 \\ {x}^{2} + 3x - 18 = 0$
$D= b^{2} -4ac=9-4\cdot (-18)=9+72=81\\x_{1,2} =\frac{-b \pm \sqrt{D}}{2a}\\x_1 = \frac{-3+9}{2}=3\\x_2 =\frac{-3-9}{2}=-6\\y_1 = 3\cdot 3 - 4 =5\\y_2 = 3 \cdot (-6) - 4 =-22$
Ответ: (3;5), (-6;-22)