Układ równań z 4 niewiadomymi

[Arek Maciejak]

Znajdż jedno z możliwych rozwiązań układu
x + s = 2(y + z )
{y + s = 3(x + z )
z + s = 5(x + y )

[agulka1987]

Znajdż jedno z możliwych rozwiązań układu
\(\displaystyle{ \begin{cases} x + s = 2(y + z )\\y + s = 3(x + z )\\z + s = 5(x + y )\end{cases}}\)

\(\displaystyle{ \begin{cases} x=2y+2z-s \\y+s=3(2y+2z-s)+3z\\z+s=5(2y+2z-s)+5y\end{cases}}\)
\(\displaystyle{ \begin{cases} x=2y+2z-s \\-5y=9z-4s \\ -9z=15y-6s \end{cases}}\)
\(\displaystyle{ \begin{cases} x=2y+2z-s \\y=\frac{4}{5}s - \frac{9}{5}z\\ z= \frac{2}{3}s - \frac{5}{3}y \end{cases}}\)
\(\displaystyle{ z= \frac{2}{3}s - \frac{5}{3}(\frac{4}{5}s - \frac{9}{5}z)}\)
\(\displaystyle{ -2z=- \frac{2}{3}s}\)
\(\displaystyle{ z= -\frac{2}{3}s * (-\frac{1}{2})=\frac{1}{3}s}\)

\(\displaystyle{ y=\frac{4}{5}s - \frac{9}{5}*\frac{1}{3}s = \frac {1}{5}s}\)

\(\displaystyle{ x=2* \frac{1}{5}s + 2*\frac{1}{3}s - s}\)
\(\displaystyle{ x=\frac{2}{5}s + \frac{2}{3}s -s= \frac{1}{15}s}\)

\(\displaystyle{ \begin{cases} x= \frac{1}{15}s\\y= \frac{1}{5}s\\z= \frac{1}{3}s \end{cases}}\)