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

$\frac{3}{4}x+\frac{1}{2}y=\frac{11}{12}$

$\frac{1}{2}x-\frac{1}{4}y=\frac{1}{8}$

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

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

Substitute $\frac{3}{4}$ in place of a, $\frac{1}{2}$ in place of c, $\frac{1}{2}$ in place of b, $-\frac{1}{4}$ in place of d, $\frac{11}{12}$ in place of e and role="math" localid="1647242577492" $\frac{1}{8}$ in place of f into role="math" localid="1647242572602" $x=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of x.

$\begin{array}{c}x=\frac{\left|\begin{array}{cc}\frac{11}{12}& \frac{1}{2}\\ \frac{1}{8}& -\frac{1}{4}\end{array}\right|}{\left|\begin{array}{cc}\frac{3}{4}& \frac{1}{2}\\ -\frac{1}{4}& -\frac{1}{4}\end{array}\right|}\\ =\frac{\left(\frac{11}{12}\right)-\left(-\frac{1}{4}\right)-\left(\frac{1}{2}\right)\cdot \left(\frac{1}{8}\right)}{\left(\frac{3}{4}\right)\cdot \left(-\frac{1}{4}\right)-\left(\frac{1}{2}\right)\cdot \frac{1}{2}}\\ =\frac{-\frac{7}{24}}{-\frac{7}{16}}\\ =\frac{2}{3}\end{array}$

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

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

Substitute $\frac{3}{4}$ in place of a, $\frac{1}{2}$ in place of c, $\frac{1}{2}$ in place of b, $-\frac{1}{4}$ in place of d, $\frac{11}{12}$ in place of e and role="math" localid="1647242799185" $\frac{1}{8}$ in place of f into role="math" localid="1647242788103" $y=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of y.

$\begin{array}{c}x=\frac{\left|\begin{array}{cc}\frac{3}{4}& \frac{11}{12}\\ \frac{1}{2}& \frac{1}{8}\end{array}\right|}{\left|\begin{array}{cc}\frac{3}{4}& \frac{1}{2}\\ -\frac{1}{4}& -\frac{1}{4}\end{array}\right|}\\ =\frac{\left(\frac{3}{4}\right)\cdot \left(\frac{1}{8}\right)-\left(\frac{11}{12}\right)\cdot \left(\frac{1}{2}\right)}{\left(\frac{3}{4}\right)\cdot \left(-\frac{1}{4}\right)-\left(\frac{1}{2}\right)\cdot \frac{1}{2}}\\ =\frac{-\frac{35}{96}}{-\frac{7}{16}}\\ =\frac{5}{6}\end{array}$

The obtained value of x is $\frac{2}{3}$ and y is $\frac{5}{6}$ so the solution is $\left(\frac{2}{3},\frac{5}{6}\right)$.