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

$\frac{1}{3}r+\frac{2}{5}s=5$

$\frac{2}{3}r-\frac{1}{2}s=-3$

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 $\frac{1}{3}$ in place of a, $\frac{2}{3}$ in place of c, $\frac{2}{5}$ in place of b, $-\frac{1}{2}$ in place of d, 5 in place of e and role="math" localid="1647241880375" $-3$ in place of f into role="math" localid="1647241874186" $r=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of r.

$\begin{array}{c}r=\frac{\left|\begin{array}{cc}5& \frac{2}{5}\\ -3& -\frac{1}{2}\end{array}\right|}{\left|\begin{array}{cc}\frac{1}{3}& \frac{2}{5}\\ \frac{2}{3}& -\frac{1}{2}\end{array}\right|}\\ =\frac{5\cdot \left(-\frac{1}{2}\right)-\left(\frac{2}{5}\right)\cdot \left(-3\right)}{\frac{1}{3}\cdot \left(-\frac{1}{2}\right)-\left(\frac{2}{5}\right)\cdot \left(\frac{2}{3}\right)}\\ =\frac{-\frac{13}{10}}{-\frac{13}{30}}\\ =3\end{array}$

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 $\frac{1}{3}$ in place of a, $\frac{2}{3}$ in place of c, $\frac{2}{5}$ in place of b, $-\frac{1}{2}$ in place of d, 5 in place of e and role="math" localid="1647242064262" $-3$ in place of f intorole="math" localid="1647242059308" $s=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of s.

$\begin{array}{c}s=\frac{\left|\begin{array}{cc}\frac{1}{3}& 5\\ \frac{2}{3}& -3\end{array}\right|}{\left|\begin{array}{cc}\frac{1}{3}& \frac{2}{5}\\ \frac{2}{3}& -\frac{1}{2}\end{array}\right|}\\ =\frac{\frac{1}{3}\cdot \left(-3\right)-5\cdot \frac{2}{3}}{\frac{1}{3}\cdot \left(-\frac{1}{2}\right)-\left(\frac{2}{5}\right)\cdot \left(\frac{2}{3}\right)}\\ =\frac{-\frac{13}{3}}{-\frac{13}{30}}\\ =10\end{array}$

The obtained value of r is 3 and s is 10 so the solution is $\left(3,10\right)$.