Question

Решить систему уравнений способом подстановки ​

Answer 1

Ответ:

$\displaystyle (-2;2),\ (2;2)$

Объяснение:

$\displaystyle \begin{cases}2x^2-y^2=4\\x^2+3y^2-5x=6\end{cases}\\\\\\\begin{cases}y^2=2x^2-4\\x^2+3y^2-5x=6\end{cases}\\\\\\\begin{cases}y^2=2x^2-4\\x^2+3(2x^2-4)-5x-6=0\end{cases}\\\\\\\begin{cases}y^2=2x^2-4\\x^2+6x^2-12-5x-6=0\end{cases}\\\\\\\begin{cases}y^2=2x^2-4\\7x^2-5x-18=0\end{cases}$

$\displaystyle 7x^2-5x-18=0\\\\a=7 ,\ \ b=-5 ,\ \ c=-18\\\\ D = b^2 - 4ac = ( - 5)^2 - 4\cdot7\cdot( - 18) = 25 + 504 = 529\\\\\sqrt{D} =\sqrt{529} = 23\\\\ x_1=\frac{-b-\sqrt{D}}{2a}=\frac{5-23}{2\cdot7}=\frac{-18 }{14 }=-\frac{9}{7}=-1\frac{2}{7} \\\\ x_2=\frac{-b+\sqrt{D}}{2a}=\frac{5+23}{2\cdot7}=\frac{28}{14}=2$

$\displaystyle y^2=2x^2-4\\\\y^2=2\cdot (-\frac{9}{7})^2-4=2\cdot \frac{81}{49}-4=\frac{162}{49}-\frac{196}{49}=-\frac{39}{49} < 0\\\\y^2=2\cdot 2^2-4=2\cdot 4-4=8-4=4\\\\y=-2\ \ \ \ \ \ \ \ \ \ \ y=2$

$\displaystyle (-2;2),\ (2;2)$