Question

Решите систему x+y=2 2x^2+xy+y^2=16

Answer 1

${x + y = 2$

${2 {x}^{2} + xy + {y}^{2} = 16$

${x = 2 - y$

${2( 2 - y)^{2} + (2 - y) y + {y}^{2} = 16$

1. Решаем квадратное уравнение.

$2(2 - y)^{2} + (2 - y) \: y + {y}^{2} = 16 \\ \\ 2( 4 - 4y + {y}^{2} ) + 2y - {y}^{2} + {y}^{2} = 16 \\ \\ 8 - 8y + 2 {y}^{2} + 2y = 16 \\ \\ 8 - 6y + 2 {y}^{2} = 16 \\ \\ 2 {y}^{2} - 6y + 8 - 16 = 0 \\ \\ 2 {y}^{2} - 6y - 8 = 0 \\ \\ d = b^{2} - 4ac \\ d = ( - 6) ^{2} - 4 \times 2 \times ( - 8) = 36 - 8 \times ( - 8) = 36 + 64 = 100 \\ \sqrt{ d} = \sqrt{100} = 10 \\ \\ x = \frac{ - b + \sqrt{d} }{2a} \\ \\ y 1= \frac{6 - 10}{2 \times 2} = - \: \frac{4}{4} = - 1 \\ \\ y2 = \frac{6 + 10}{2 \times 2} = \frac{16}{4} = 4 \\$

$otvet : y1 = - 1. \: y2 = 4$

2. Решаем систему уравнение.

1. [х=2-у

[ у1= - 1

[ х= 2- (-1)

[у1= -1

[х= 2+1=3

[у1= - 1

2. [ х= 2- 4

[ у2=4

[х2= - 2

[ у2= 4

ответ: х= 3 и у= - 1; х2= - 2 и у2= 4

Вот