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 $x_1=\frac{\begin{vmatrix}
d_1 &b_1 &c_1\\d_2&b_2 & c_2\\d_3&b_3&c_3
\end{vmatrix}}{\begin{vmatrix}
a_1 &b_1&c_1\\a_2&b_2 &c_2\\a_3&b_3&c_3
\end{vmatrix}}=\frac{\begin{vmatrix}
5 & 1 & 3\\7&-1&5\\2&3&1
\end{vmatrix}}{\begin{vmatrix}
4&1&3\\2&-1&5\\2&3&1
\end{vmatrix}}=\frac{5(-1-15)-1(7-10)+3(21+2)}{4(-1-15)-1(2-10)+3(6+2)}=\frac{-8}{-32}=\frac{1}{4}$
$x_2=\frac{\begin{vmatrix}
a_1 &d_1 &c_1\\a_2&d_2& c_2\\a_3&d_3&c_3
\end{vmatrix}}{\begin{vmatrix}
a_1 &b_1&c_1\\a_2&b_2 &c_2\\a_3&b_3&c_3
\end{vmatrix}}=\frac{\begin{vmatrix}
4 & 5 & 3\\2&7&5\\2&2&1
\end{vmatrix}}{\begin{vmatrix}
4&1&3\\2&-1&5\\2&3&1
\end{vmatrix}}=\frac{4(7-10)-5(2-10)+3(4-14)}{4(-1-15)-1(2-10)+3(6+2)}=\frac{-2}{-32}=\frac{1}{16}$
and $x_3=\frac{\begin{vmatrix}
d_1 &b_1 &c_1\\d_2&b_2 & c_2\\d_3&b_3&c_3
\end{vmatrix}}{\begin{vmatrix}
a_1 &b_1&c_1\\a_2&b_2 &c_2\\a_3&b_3&c_3
\end{vmatrix}}=\frac{\begin{vmatrix}
4& 1 & 5\\2&-1&7\\2&3&2
\end{vmatrix}}{\begin{vmatrix}
4&1&3\\2&-1&5\\2&3&1
\end{vmatrix}}=\frac{4(-2-21)-1(4-14)+5(6+2)}{4(-1-15)-1(2-10)+3(6+2)}=\frac{-42}{-32}=\frac{21}{16}$
