Answer
$(\frac{860}{41}, -\frac{540}{41}, \frac{10}{41})$
Work Step by Step
Step 1. Define the following matrices based on the given system of equations:
$\begin{array} \\D=\\ \end{array}
\begin{bmatrix} 3&4&-1\\ 1&0&-4\\2&1&5\end{bmatrix},
\begin{array} \\D_x=\\ \end{array}
\begin{bmatrix} 10&4&-1\\20&0&-4\\30&1&5\end{bmatrix},
\begin{array} \\D_y=\\ \end{array}
\begin{bmatrix} 3&10&-1\\ 1&20&-4\\2&30&5\end{bmatrix},
\begin{array} \\D_z=\\ \end{array}
\begin{bmatrix} 3&4&10\\ 1&0&20\\2&1&30\end{bmatrix}$
Step 2. Calculate the determinants of the above matrices (use column 2 expansions):
$|D|=-4(5+8)-1(-12+1)=-41$, $|D_x|=-4(100+120)-1(-40+20)=-860$, $|D_y|=-10(5+8)+20(15+2)-30(-12+1)=540$, $|D_z|=-4(30-40)-1(60-10)=-10$
Step 3. Use the Cramer's Rule:
$x=\frac{|D_x|}{|D|}=\frac{860}{41}$, $y=\frac{|D_y|}{|D|}=-\frac{540}{41}$, $z=\frac{|D_z|}{|D|}=\frac{10}{41}$
Step 4. Conclusion: the solution to the system is $(\frac{860}{41}, -\frac{540}{41}, \frac{10}{41})$
