91Ó°ÊÓ

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 $1.5$ in place of a, $2.2$ in place of c, $-0.7$ in place of b, $-0.6$ in place of d, $0.5$ in place of e and role="math" localid="1647193225857" $-7.4$ in place of f into role="math" localid="1647193212588" $m=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of m.

$\begin{array}{c}m=\frac{\left|\begin{array}{cc}0.5& −0.7\\ −7.4& −0.6\end{array}\right|}{\left|\begin{array}{cc}1.5& −0.7\\ 2.2& −0.6\end{array}\right|}\\ =\frac{0.5â‹\dots \left(−0.6\right)−\left(−7.4\right)â‹\dots \left(−0.7\right)}{1â‹\dots 5â‹\dots \left(−0.6\right)−2.2â‹\dots \left(−0.7\right)}\\ =\frac{−5.48}{0.64}\\ =−8.5625\end{array}$

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 $1.5$ in place of a, $2.2$ in place of c, $-0.7$ in place of b, $-0.6$ in place of d, role="math" localid="1647193438046" $0.5$ in place of e and role="math" localid="1647193431538" $-7.4$ in place of f into role="math" localid="1647193422851" $n=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of n.

$\begin{array}{c}m=\frac{\left|\begin{array}{cc}1.5& 0.5\\ 2.2& −7.4\end{array}\right|}{\left|\begin{array}{cc}1.5& −0.7\\ 2.2& −0.6\end{array}\right|}\\ =\frac{1.5â‹\dots \left(−7.4\right)−2.2â‹\dots 0.5}{1â‹\dots 5â‹\dots \left(−0.6\right)−2.2â‹\dots \left(−0.7\right)}\\ =\frac{−12.2}{0.64}\\ =−19.0625\end{array}$

The obtained value of m is $-8.5625$ and n is $-19.0625$ so the solution is $\left(-8.5625,-19.0625\right)$.