Answer:
[tex]r=4\\ \\s=3\\ \\t=-1[/tex]
Step-by-step explanation:
Given the system of three linear equations:
[tex]\left\{\begin{array}{l}-6r+5s+2t=-11\\ \\-2r+s+4t=-9\\ \\4r-5s+5t=-4\end{array}\right.[/tex]
Cheange the positions of the first and the second equations:
[tex]\left\{\begin{array}{l}-2r+s+4t=-9\\ \\-6r+5s+2t=-11\\ \\4r-5s+5t=-4\end{array}\right.[/tex]
Multiply the first equation by -3 and add it to the second equation and then multiply the first equation by 2 and add it to the third equation:
[tex]\left\{\begin{array}{r}-2r+s+4t=-9\\ \\2s-10t=16\\ \\-3s+13t=-22\end{array}\right.[/tex]
Multiply the second equation by 3, the third equation by 2 and add them:
[tex]\left\{\begin{array}{r}-2r+s+4t=-9\\ \\2s-10t=16\\ \\-4t=4\end{array}\right.[/tex]
From the third equation,
[tex]t=-1[/tex]
Substitute it into the second equation:
[tex]2s-10\cdot (-1)=16\\ \\2s+10=16\\ \\2s=6\\ \\s=3[/tex]
Substitute [tex]t=-1[/tex] and [tex]s=3[/tex] into the first equation:
[tex]-2r+3+4\cdot (-1)=-9\\ \\-2r+3-4=-9\\ \\-2r-1=-9\\ \\-2r=-9+1\\ \\-2r=-8\\ \\2r=8\\ \\r=4[/tex]
Hence,
[tex]r=4\\ \\s=3\\ \\t=-1[/tex]
