Use Cramer’s Rule to solve each system of equations.

$2m-4n=-1$

$3n-4m=-5$

Apply the Cramer’s rule for the value of m.

$m=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$

Substitute 2 in place of a, $-4$ in place of c, $-4$ in place of b, 3 in place of d, $-1$ in place of e and role="math" localid="1647191286819" $-5$ in place of f into role="math" localid="1647191280200" $m=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of m.

$\begin{array}{c}m=\frac{\left|\begin{array}{cc}-1& -4\\ -5& 3\end{array}\right|}{\left|\begin{array}{cc}2& -4\\ -4& 3\end{array}\right|}\\ =\frac{\left(-1\right)\cdot 3-\left(-4\right)\cdot \left(-5\right)}{2\cdot 3-\left(-4\right)\cdot \left(-4\right)}\\ =\frac{-23}{-10}\\ =\frac{23}{10}\end{array}$

Apply the Cramer’s rule for the value of n.

$n=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$

Substitute 2 in place of a, $-4$ in place of c, $-4$ in place of b, 3 in place of d, $-1$ in place of e and role="math" localid="1647191472584" $-5$ in place of f into role="math" localid="1647191466954" $n=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of n.

$\begin{array}{c}n=\frac{\left|\begin{array}{cc}2& -1\\ -4& -5\end{array}\right|}{\left|\begin{array}{cc}2& -4\\ -4& 3\end{array}\right|}\\ =\frac{2\cdot \left(-5\right)-\left(-1\right)\cdot \left(-4\right)}{2\cdot 3-\left(-4\right)\cdot \left(-4\right)}\\ =\frac{-14}{-10}\\ =\frac{14}{10}\end{array}$

The obtained value of m is 5 and n is 1 so the solution is $\left(\frac{23}{10},\frac{7}{5}\right)$.