Решите систему уравнений (х-6)(у-5)=0 у-2/х+у-8=3. Created by vikakochubey. algebra-ru

х+у-8=3...

vikakochubey @vikakochubey

August 2022 1 8 Report

Решите систему уравнений
(х-6)(у-5)=0
у-2/х+у-8=3

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lilyatomach

Verified answer

Ответ:

(4;5)

Объяснение:

ОДЗ : $x+y-8\neq 0;x\neq 8-y.$

$\left \{ \begin{array}{lcl} {(x-6)({y-5)=0,} \\ {\frac{y-2}{x+y-8} =3;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{(x-6)(y-5)=0,} \\ {y-2=3x+3y-24;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{(x-6)(y-5)=0,} \\ {3x+2y=22;}} \end{array} \right.$

$\Leftrightarrow\left \{ \begin{array}{lcl} {\left [ \begin{array}{lcl} {{x-6=0,} \\ {y-5=0,}} \end{array} \right.{} \\ {3x+2y=22;}} \end{array} \right.\Leftrightarrow \left [ \begin{array}{lcl} {{\left \{ \begin{array}{lcl} {{x=6} \\ {3*6+2y=22;}} \end{array} \right.} \\ \left \{ \begin{array}{lcl} {{y=5,} \\ {3x+2*5=22;}} \end{array} \right.{}} \end{array} \right.\Leftrightarrowt \Leftrightarrow$

$\Leftrightarrow\left [ \begin{array}{lcl} {{\left \{ \begin{array}{lcl} {{x=6,} \\ {y=2;}} \end{array} \right.} \\ {\left \{ \begin{array}{lcl} {{y=5,} \\ {x=4.}} \end{array} \right.}} \end{array} \right.$

если x=6,y=2 , то знаменатель дроби обращается в нуль. Значит эта пара не удовлетворяет ОДЗ и не является решением.

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