Question

Решите способом подстановки систему уравнений: г)

Answer 1

Ответ:

$80a)\ \ \left\{\begin{array}{l}x+y=2\\x^2-y^2=100\end{array}\right\ \ \left\{\begin{array}{l}x+y=2\\(x-y)(x+y)=100\end{array}\right\ \ \left\{\begin{array}{l}x+y=2\\(x-y)\cdot 2=100\end{array}\right\\\\\\\left\{\begin{array}{l}x+y=2\\x-y=50\end{array}\right\ \oplus\ \ominus\ \left\{\begin{array}{l}2x=52\\2y=-48\end{array}\right\ \ \left\{\begin{array}{l}x=26\\y=-24\end{array}\right\ \ \ Otvet:\ (26;-24)\ .$

$b)\ \ \left\{\begin{array}{l}x+y=-5\\x^2-y^2=5\end{array}\right\ \ \left\{\begin{array}{l}x+y=-5\\(x-y)(x+y)=5\end{array}\right\ \ \left\{\begin{array}{l}x+y=5\\-5\, (x-y)=5\end{array}\right\\\\\\\left\{\begin{array}{l}x+y=-5\\x-y=-1\end{array}\right\ \oplus \ominus \ \left\{\begin{array}{l}2x=-6\\2y=-4\end{array}\right\ \ \left\{\begin{array}{l}x=-3\\y=-2\end{array}\right\ \ \ Otvet:\ (-3;-2)\ .$

$c)\ \ \left\{\begin{array}{l}y-3x=0\\x^2+y^2=40\end{array}\right\ \ \left\{\begin{array}{l}y=3x\\x^2+(3x)^2=40\end{array}\right\ \ \left\{\begin{array}{l}y=3x\\10x^2=40\end{array}\right\\\\\\\left\{\begin{array}{l}y=3x\\x^2=4\end{array}\righty\ \ \left\{\begin{array}{l}y=3x\\x=\pm 2\end{array}\right$

$1.\ \left\{\begin{array}{l}y=3x\\x=-2\end{array}\right\ \ \left\{\begin{array}{l}y=-6\\x=-2\end{array}\right\\\\\\2.\ \ \left\{\begin{array}{l}y=3x\\x=2\end{array}\right\ \ \left\{\begin{array}{l}y=6\\x=2\end{array}\right\\\\\\Otvet:\ \ (-2;-6)\ ,\ \ (2;6)\ .$

$d)\ \ \left\{\begin{array}{l}2x+y=0\\xy=2\end{array}\right\ \ \left\{\begin{array}{l}y=-2x\\x(-2x)=2\end{array}\right\ \ \left\{\begin{array}{l}y=-2x\\-2x^2=2\end{array}\right\ \ \left\{\begin{array}{l}y=-2x\\x^2=-1\end{array}\right\\\\\\Otvet:\ \ \varnothing \ ,\ tak\ kak\ \ x^2\geq 0\ \ i\ \ x^2\ne -1\ .$

P.S.  Квадрат любого числа неотрицателен, поэтому он не может равняться (-1), значит система не имеет решений .