91Ó°ÊÓ

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

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

Substitute $\frac{1}{2}$ in place of a, $\frac{3}{5}$ in place of c, $-\frac{2}{3}$ in place of b, $\frac{4}{5}$ in place of d, $2\frac{1}{3}$ in place of e and role="math" localid="1647356974381" $-10$ in place of f into role="math" localid="1647356967875" $r=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of r.

$\begin{array}{c}r=\frac{\left|\begin{array}{cc}2\frac{1}{3}& −\frac{2}{3}\\ −10& \frac{4}{5}\end{array}\right|}{\left|\begin{array}{cc}\frac{1}{2}& −\frac{2}{3}\\ \frac{3}{5}& \frac{4}{5}\end{array}\right|}\\ =\frac{\frac{7}{3}â‹\dots \frac{4}{5}−\left(−\frac{2}{3}\right)â‹\dots \left(−10\right)}{\frac{1}{2}â‹\dots \frac{4}{5}−\left(−\frac{2}{3}\right)â‹\dots \frac{3}{5}}\\ =\frac{−\frac{24}{5}}{\frac{4}{5}}\\ =−6\end{array}$

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

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

Substitute $\frac{1}{2}$ in place of a, $\frac{3}{5}$ in place of c, $-\frac{2}{3}$ in place of b, $\frac{4}{5}$ in place of d, $2\frac{1}{3}$ in place of e and role="math" localid="1647357186062" $-10$ in place of f intorole="math" localid="1647357180824" $s=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of s.

$\begin{array}{c}s=\frac{\left|\begin{array}{cc}\frac{1}{2}& 2\frac{1}{3}\\ \frac{3}{5}& −10\end{array}\right|}{\left|\begin{array}{cc}\frac{1}{2}& −\frac{2}{3}\\ \frac{3}{5}& \frac{4}{5}\end{array}\right|}\\ =\frac{\frac{1}{2}â‹\dots \left(−10\right)−\left(\frac{7}{3}\right)â‹\dots \left(\frac{3}{5}\right)}{\frac{1}{2}â‹\dots \frac{4}{5}−\left(−\frac{2}{3}\right)â‹\dots \frac{3}{5}}\\ =\frac{−\frac{32}{5}}{\frac{4}{5}}\\ =−8\end{array}$

The obtained value of r is $-6$ and s is $-8$ so the solution is $\left(-6,-8\right)$.