Question

Решите систему из двух выражений
х² + 2х = у² - 1
у² - 4у = х² - 5

Answer 1

$\left\{\begin{array}{ccc}x^{2} +2x=y^{2} -1 \\y^{2}-4y=x^{2}-5\end{array}\right \\\\\\+\left[\begin{array}{ccc}x^{2}-y^{2}+2x=-1\\-x^{2}+y^{2} -4y=-5 \end{array}\right\\-------------\\2x-4y=-6\\\\x-2y=-3\\\\x=2y-3\\\\\left\{\begin{array}{ccc}x=2y-3\\y^{2} -4y=x^{2} -5 \end{array}\right \\\\\\\left\{\begin{array}{ccc}x=2y-3\\y^{2} -4y=(2y-3)^{2} -5 \end{array}\right\\\\\\\left\{\begin{array}{ccc}x=2y-3\\y^{2} -4y=4y^{2}-12y+9 -5 \end{array}\right$

$\left\{\begin{array}{ccc}x=2y-3\\3y^{2}-8y+4=0 \end{array}\right\\\\3y^{2}-8y+4=0\\\\D=(-8)^{2}-4*3*4=64-48=16=4^{2}\\\\y_{1}=\dfrac{8-4}{6}=\frac{2}{3} \\\\y_{2}=\dfrac{8+4}{6}=2 \\\\x_{1}=2*\frac{2}{3} -3=1\dfrac{1}{3}-3=-1\dfrac{2}{3}\\\\x_{2}=2*2-3=1\\\\Otvet:\boxed{\Big(-1\dfrac{2}{3} \ ; \ \frac{2}{3}\Big) \ , \ \Big(1 \ ; \ 2\Big)}$