Решите систему уравнения : (х-4)(у-6)=0,
у-4 / х + у -8 =2

Ответ:

{(3; 6)}

Объяснение:

\displaystyle \tt \left \{\begin{array}{ccc}x+y-8\neq 0 \\(x-4) \cdot (y-6)=0\\\dfrac{y-4}{x+y-8}=2 \end{array}\right \Leftrightarrow \left \{\begin{array}{ccc}x+y-8\neq 0 \\(x-4) \cdot (y-6)=0\\ y-4=2 \cdot (x+y-8) \end{array}\right \Leftrightarrow

\displaystyle \tt \Leftrightarrow \left \{\begin{array}{ccc}x+y-8\neq 0 \\(x-4) \cdot (y-6)=0\\ y-4=2 \cdot x+2 \cdot y-16 \end{array}\right \Leftrightarrow \left \{\begin{array}{ccc}x+y-8\neq 0 \\(x-4) \cdot (y-6)=0\\ 2 \cdot x+y=12 \end{array}\right

(x-4)·(y-6)=0 ⇔ x-4=0 или y-6=0 ⇔ x=4 или y=6.

1) x=4

\displaystyle \tt \left \{\begin{array}{ccc}x+y-8\neq 0 \\ x=4\\ 2 \cdot x+y=12 \end{array}\right \Leftrightarrow \left \{\begin{array}{ccc} 4+y-8\neq 0 \\ x=4 \\ 2 \cdot 4+y=12 \end{array}\right \Leftrightarrow

\displaystyle \tt \Leftrightarrow \left \{\begin{array}{ccc}y-4\neq 0 \\ x=4 \\ y=12-8=4 \end{array}\right \Rightarrow y \in \varnothing.

2) y=6

\displaystyle \tt \left \{\begin{array}{ccc}x+y-8\neq 0 \\ y=6\\ 2 \cdot x+y=12 \end{array}\right \Leftrightarrow \left \{\begin{array}{ccc} x+6-8\neq 0 \\ y=6 \\ 2 \cdot x+6=12 \end{array}\right \Leftrightarrow

\displaystyle \tt \Leftrightarrow \left \{\begin{array}{ccc} x \neq 2 \\ y=6 \\ x=3 \end{array}\right \Rightarrow (x; \; y) = (3; \; 6).