Question

решите систему уравнений

x² + xy = 6
xy + y^2=10

Answer 1

Ответ:

Решить систему уравнений    $\left\{\begin{array}{l}\bf x^2+xy=6\\\bf xy+y^2=10\end{array}\right$   .

Сложим оба уравнения и применим формулу квадрата суммы .

$\left\{\begin{array}{l}\bf x^2+xy=6\\\bf x^2+2xy+y^2=16\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+xy=6\\\bf (x+y)^2=16\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+xy=6\\\bf x+y=\pm 4\end{array}\right$  

Рассмотрим два случая .

$\bf a)\ \ \left\{\begin{array}{l}\bf x^2+xy=6\\\bf x+y=4\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+xy=6\\\bf y=4-x\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+x\, (4-x)=6\\\bf y=4-x\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+4x-x^2=6\\\bf y=4-x\end{array}\right$

$\left\{\begin{array}{l}\bf 4x=6\\\bf y=4-x\end{array}\right\ \ \left\{\begin{array}{l}\bf x=1,5\\\bf y=4-1,5\end{array}\right\ \ \left\{\begin{array}{l}\bf x=1,5\\\bf y=2,5\end{array}\right\ \ \ \Rightarrow \ \ \ \bf (\ 1,5\ ;\ 2,5\ )$

$\bf b)\ \ \left\{\begin{array}{l}\bf x^2+xy=6\\\bf x+y=-4\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+xy=6\\\bf y=-4-x\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+x\, (-4-x)=6\\\bf y=-4-x\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2-4x-x^2=6\\\bf y=-4-x\end{array}\right$

$\left\{\begin{array}{l}\bf -4x=6\\\bf y=-4-x\end{array}\right\ \ \left\{\begin{array}{l}\bf x=-1,5\\\bf y=-4-(-1,5)\end{array}\right\ \ \left\{\begin{array}{l}\bf x=-1,5\\\bf y=-2,5\end{array}\right\ \ \ \Rightarrow \ \ \ \bf (\, -1,5\ ;\, -2,5\ )$  

       

Ответ:  $\boldsymbol{(\ 1,5\ ;\ 2,5\ )\ ,\ (\, -1,5\ ;\, -2,5\ )}$  .