Question

Помогите решить СрОчНо!

Answer 1

$1)\ \ \left\{\begin{array}{l}4x-6y=26\ |:2\\5x+3y=1\end{array}\right\ \ \left\{\begin{array}{l}2x-3y=13\\5x+3y=1\end{array}\right\ \oplus \ \left\{\begin{array}{l}7x=14\\3y=1-5x\end{array}\right\\\\\\\left\{\begin{array}{l}x=2\\3y=1-10\end{array}\right\ \ \left\{\begin{array}{l}x=2\\3y=-9\end{array}\right\ \ \left\{\begin{array}{l}x=2\\y=-3\end{array}\right\ \ \ \ Otvet:\ (2;-3)$

$2)\ \ \left\{\begin{array}{l}x-y=4\\x^2+y^2=10\end{array}\right\ \ \left\{\begin{array}{l}x-y=4\\(x^2-2xy+y^2)+2xy=10\end{array}\right\ \ \left\{\begin{array}{l}x-y=4\\(x-y)^2+2xy=10\end{array}\right\\\\\\\left\{\begin{array}{l}x-y=4\\16+2xy=10\end{array}\right\ \ \left\{\begin{array}{l}x-y=4\\2xy=-6\end{array}\right\ \ \left\{\begin{array}{l}x-y=4\\xy=-3\end{array}\right\ \ \left\{\begin{array}{l}y=x-4\\x(x-4)=-3\end{array}\right$

$\left\{\begin{array}{l}y=x-4\\x^2-4x+3=0\end{array}\right\ \ \left\{\begin{array}{l}y_1=-3\ ,\ y_2=-1\\x_1=1\ ,\ x_2=3\end{array}\right\\\\\\Otvet:\ \ (1;-3)\ ,\ (3;-1)\ .$

$3)\ \ \left\{\begin{array}{l}\dfrac{x+2y}{4}-\dfrac{x-2y}{2}=\dfrac{7-2y}{3}-x+1\\3x-2y=8\end{array}\right\\\\\\\left\{\begin{array}{l}3(x+2y)-6(x-2y)=4(7-2y)-12x+12\\3x-2y=8\end{array}\right\\\\\\\left\{\begin{array}{l}3x+6y-6x+12y=28-8y-12x+12\\3x-2y=8\end{array}\right$

$\left\{\begin{array}{l}9x+26y=40\\3x-2y=8\ |\cdot (-3)\end{array}\right\ \oplus\ \left\{\begin{array}{l}9x=40-26y\\32y=16\end{array}\right\ \ \left\{\begin{array}{l}9x=27\\y=0,5\end{array}\right\\\\\\\left\{\begin{array}{l}x=3\\y=0,5\end{array}\right\ \ \ \ \ Otvet:\ \ (\ 3\ ;\ 0,5\ )\ .$