§8. Неравенства с двумя переменными и их системы — Дополнительные упражнения к параграфу 7 — 493 — стр. 145

$\{\begin{array}{l}x-y=4\\ (x-1)(y+1)=2xy+3\end{array}$

$\{\begin{array}{l}y=x-4\\ (x-1)(x-4+1)=2x(x-4)+3\end{array}$

$\{\begin{array}{l}y=x-4\\ (x-1)(x-3)=2x^2-8x+3\end{array}$

$\{\begin{array}{l}y=x-4\\ x^2-4x+3-2x^2+8x-3=0\end{array}$

$$\begin{gathered} x^2-4x=0 \\ x(x-4)=0 \\ {x}_1=0 \\ {x}_2=4 \end{gathered}$$

$\{\begin{array}{l}x=0\\ y=-4\end{array}$ или $\{\begin{array}{l}x=4\\ y=0\end{array}$

Ответ: $(0;-4),(4;0)$.

$\{\begin{array}{l}y-x=1\\ (2y+1)(x-1)=xy+1\end{array}$

$\{\begin{array}{l}y=1+x\\ (2+2x+1)(x-1)=x(1+x)+1\end{array}$

$\{\begin{array}{l}y=1+x\\ 2x^2+3x-2x-3-x-x^2-1=0\end{array}$

$$\begin{gathered} x^2-4=0 \\ x=\pm 2 \end{gathered}$$

$\{\begin{array}{l}x=2\\ y=3\end{array}$ или $\{\begin{array}{l}x=-2\\ y=-1\end{array}$

Ответ: $(2;3),(-2;-1)$.

$\{\begin{array}{l}2x-y=5\\ (x+1)(y+4)=2xy-1\end{array}$

$\{\begin{array}{l}y=2x-5\\ (x+1)(2x-1)-2x(2x-5)+1=0\end{array}$

$\{\begin{array}{l}y=2x-5\\ 2x^2+2x-x-1-4x^2+10x+1=0\end{array}$

$\{\begin{array}{l}y=2x-5\\ -2x^2+11x=0\end{array}$

$$\begin{gathered} 2x^2-11x=0 \\ x(2x-11)=0 \end{gathered}$$

$x_1=0$

$x_2=5,5$

$\{\begin{array}{l}x=0\\ y=-5\end{array}$ или $\{\begin{array}{l}x=5,5\\ y=6\end{array}$

Ответ: $(0;-5),(5,5;6)$.

$\{\begin{array}{l}x+y=1\\ (x-1)(y+5)=y^2-12\end{array}$

$\{\begin{array}{l}y=1-x\\ (x-1)(1-x+5)=(1-x{)}^2-12\end{array}$

$\{\begin{array}{l}y=1-x\\ 6x-6-x^2+x=1-2x+x^2-12\end{array}$

$\{\begin{array}{l}y=1-x\\ 1-2x+x^2-12-6x+6+x^2-x=0\end{array}$

$$\begin{gathered} 2x^2-9x-5=0 \\ {x}_{1,2}=\frac{9\pm \sqrt{81+40}}{4} \\ {x}_1=5, \\ {x}_2=-\frac{1}{2} \end{gathered}$$

$\{\begin{array}{l}x=5\\ y=-4\end{array}$ или $\{\begin{array}{l}x=-\frac{1}{2}\\ y=1,5\end{array}$

Ответ: $(5;-4),(-0,5;1,5)$.