We have the following system of equations
[tex]\begin{gathered} -3x-5y+36x=76 \\ -x+7z=12 \\ 1+x+y-10z=-19 \end{gathered}[/tex]By subtracting 1 to the last equation, we have an equivalent system of equations:
[tex]\begin{gathered} -3x-5y+36x=76 \\ -x+7z=12 \\ x+y-10z=-20 \end{gathered}[/tex]and we need to find the solutions (or no solutions) of this system.
Solving by elimination method.
By adding the second and third equations, we have
[tex]y-3z=-8\ldots(iv)[/tex]so we have eliminated the variable x. Now, by multiplying by -3 the second equation
[tex]\begin{gathered} -3x-5y+36x=76 \\ 3x-21z=-36 \\ \\ \end{gathered}[/tex]and adding the result to the first one, we get
[tex]-5y+15z=40\ldots(v)[/tex]By combining equations (iv) and (v), we have the following system in 2 variables:
[tex]\begin{gathered} y-3z=-8 \\ -5y+15z=40 \end{gathered}[/tex]Lets eliminate the variable z. In this regard, by multiplying the first equation by 5, we have
[tex]\begin{gathered} 5y-15z=-40 \\ -5y+15z=40 \end{gathered}[/tex]and by adding both equations, we get
[tex]0=0[/tex]This means that there are dependent equations. So, we can write 2 variables in terms on one of them. Therefore, the answer is option 3: Infinitely many solutions.
