Answer
$x_2=\frac{9}{4}$
Work Step by Step
We are given:
$x_1+4x_2-2x_3+x_4=2$
$2x_1+9x_2-3x_3-2x_4=5$
$x_1+5x_2+x_3-x_4=3$
$3x_1+14x_2+7x_3-2x_4=6$
Using Cramer's Rule to find $x_2$:
$x_2=\frac{\det(B_2)}{\det(A)}$
which can also be written as:
$A=\begin{bmatrix}
1 &4&-2&1\\
2 & 9&-3&-2\\
1&5&1&-1 \\
3&14&7&-2
\end{bmatrix} \rightarrow \det(A)=1.15-2.49+1.205-3.46=-16$
$B_2=\begin{bmatrix}
1 & 2&-2&1\\
2 & 4&-3&-2\\
1&3&1&-1 \\
3&6&7&-2
\end{bmatrix} \rightarrow \det(B_2)=1.(-5)-2.25+1.97-3.26=-36$
Hence, $x_2=\frac{\det(B_2)}{\det(A)}=\frac{-36}{-16}=\frac{9}{4}$
