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

$2a+3b=-16$

$\frac{3}{4}a-\frac{7}{8}b=10$

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

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

Substitute 2 in place of a, $\frac{3}{4}$ in place of c, 3 in place of b, $-\frac{7}{8}$ in place of d, $-16$ in place of e and 10 in place of f into role="math" localid="1647241145390" $a=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of a.

$\begin{array}{c}a=\frac{\left|\begin{array}{cc}-16& 3\\ 10& -\frac{7}{8}\end{array}\right|}{\left|\begin{array}{cc}2& 3\\ \frac{3}{4}& -\frac{7}{8}\end{array}\right|}\\ =\frac{\left(-16\right)\cdot \left(-\frac{7}{8}\right)-3\cdot 10}{2\cdot \left(-\frac{7}{8}\right)-3\cdot \left(\frac{3}{4}\right)}\\ =\frac{-16}{-4}\\ =4\end{array}$

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

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

Substitute 2 in place of a, $\frac{3}{4}$ in place of c, 3 in place of b, $-\frac{7}{8}$ in place of d, $-16$ in place of e and 10 in place of f into role="math" localid="1647241359251" $b=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of b.

$\begin{array}{c}a=\frac{\left|\begin{array}{cc}2& -16\\ \frac{3}{4}& 10\end{array}\right|}{\left|\begin{array}{cc}2& 3\\ \frac{3}{4}& -\frac{7}{8}\end{array}\right|}\\ =\frac{2\cdot 10-\left(-16\right)\cdot \left(\frac{3}{4}\right)}{2\cdot \left(-\frac{7}{8}\right)-3\cdot \left(\frac{3}{4}\right)}\\ =\frac{32}{-4}\\ =-8\end{array}$

The obtained value of a is 4 and b is $-8$ so the solution is $\left(4,-8\right)$.