Answer
See below
Work Step by Step
According to Cramer's Rule for a $3\times 3$ system $Ax=b$ where $A=\begin{bmatrix}
a_1 &b_1 &c_1\\a_2&b_2 &c_2\\a_3&b_3&c_3
\end{bmatrix}$ and $b=\begin{bmatrix}
c_1\\c_2\\c_3
\end{bmatrix}$
we have $x_1=\frac{\begin{vmatrix}
d_1 &b_1 &c_1\\d_2&b_2 & c_2\\d_3&b_3&c_3
\end{vmatrix}}{\begin{vmatrix}
a_1 &b_1&c_1\\a_2&b_2 &c_2\\a_3&b_3&c_3
\end{vmatrix}}=\frac{\begin{vmatrix}
3 & 3 & 6\\-1&4&-7\\4&5&9
\end{vmatrix}}{\begin{vmatrix}
5&3&6\\2&4&-7\\2&5&9
\end{vmatrix}}=\frac{3(36+35)-3(-9+28)+6(-5-16)}{5(36+35)-3(18+14)+6(10-8)}=\frac{30}{271}$
$x_2=\frac{\begin{vmatrix}
a_1 &d_1 &c_1\\a_2&d_2& c_2\\a_3&d_3&c_3
\end{vmatrix}}{\begin{vmatrix}
a_1 &b_1&c_1\\a_2&b_2 &c_2\\a_3&b_3&c_3
\end{vmatrix}}=\frac{\begin{vmatrix}
5 & 3 & 6\\2&-1&-7\\2&4&9
\end{vmatrix}}{\begin{vmatrix}
5&3&6\\2&4&-7\\2&5&9
\end{vmatrix}}=\frac{5(-9+28)-3(18+14)+6(8+2)}{5(36+35)-3(18+14)+6(10-8)}=\frac{59}{271}$
and $x_3=\frac{\begin{vmatrix}
d_1 &b_1 &c_1\\d_2&b_2 & c_2\\d_3&b_3&c_3
\end{vmatrix}}{\begin{vmatrix}
a_1 &b_1&c_1\\a_2&b_2 &c_2\\a_3&b_3&c_3
\end{vmatrix}}=\frac{\begin{vmatrix}
3 & 3 & 6\\-1&4&-7\\4&5&9
\end{vmatrix}}{\begin{vmatrix}
5&3&3\\2&4&-7\\2&5&9
\end{vmatrix}}=\frac{5(16+5)-3(8+2)+3(10-8)}{5(36+35)-3(18+14)+6(10-8)}=\frac{81}{271}$
