Respuesta :
To solve the system of equations:
[tex]\begin{gathered} -3a-b-3c=-8 \\ -5a+3b+6c=-4 \\ -6a-4b+c=-20 \end{gathered}[/tex]by elimination we choose two equations and eliminate one of the variables. Choosing the first two equations and multiplying the first one by 2 we have:
[tex]\begin{gathered} -6a-2b-6c=-16 \\ -5a+3b+6c=-4 \end{gathered}[/tex]if we add the equation we get:
[tex]-11a+b=-20[/tex]Now, from the original system we choose the second and third equations and multyply the third by -6, then we have:
[tex]\begin{gathered} -5a+3b+6c=-4 \\ 36a+24b-6c=120 \end{gathered}[/tex]Adding the two equations we have:
[tex]31a+27b=116[/tex]Now that we eliminate the same variable from two sets of equations we have the new system:
[tex]\begin{gathered} -11a+b=-20 \\ 31a+27b=116 \end{gathered}[/tex]To solve this sytem we mutiply the first equation by -27, then we have:
[tex]\begin{gathered} 297a-27b=540 \\ 31a+27b=116 \end{gathered}[/tex]adding this equation we have:
[tex]\begin{gathered} 328a=656 \\ a=\frac{656}{328} \\ a=2 \end{gathered}[/tex]Once we have the value of a we plug it in the equation
[tex]-11a+b=-20[/tex]to find b:
[tex]\begin{gathered} -11(2)+b=-20 \\ -22+b=-20 \\ b=22-20 \\ b=2 \end{gathered}[/tex]Now that we have the values of a and b we plug them in the equation
[tex]-3a-b-3c=-8[/tex]to find c:
[tex]\begin{gathered} -3(2)-2-3c=-8 \\ -6-2-3c=-8 \\ -8-3c=-8 \\ 3c=-8+8 \\ 3c=0 \\ c=\frac{0}{3} \\ c=0 \end{gathered}[/tex]Therefore the solution of the system is:
[tex]\begin{gathered} a=2 \\ b=2 \\ c=0 \end{gathered}[/tex]