91Ó°ÊÓ

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 3 in place of a, 5 in place of c, 8 in place of b, $-7$ in place of d, 28 in place of e and role="math" localid="1647279188349" $-55$ in place of f intorole="math" localid="1647279134383" $x=\frac{\left|\begin{array}{cc}e& b\\ f& d\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of x.

$\begin{array}{c}x=\frac{\left|\begin{array}{cc}28& 8\\ −55& −7\end{array}\right|}{\left|\begin{array}{cc}3& 8\\ 5& −7\end{array}\right|}\\ =\frac{28â‹\dots \left(−7\right)−8â‹\dots \left(−55\right)}{3â‹\dots \left(−7\right)−8â‹\dots 5}\\ =\frac{244}{−61}\\ =−4\end{array}$

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 3 in place of a, 5 in place of c, 8 in place of b, $-7$ in place of d, 28 in place of e and localid="1647279403902" $-55$ in place of f into localid="1647279397864" $y=\frac{\left|\begin{array}{cc}a& e\\ c& f\end{array}\right|}{\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|}$and perform simplification of determinants to write the value of y.

$\begin{array}{c}y=\frac{\left|\begin{array}{cc}3& 28\\ 5& −55\end{array}\right|}{\left|\begin{array}{cc}3& 8\\ 5& −7\end{array}\right|}\\ =\frac{3â‹\dots \left(−55\right)−\left(28\right)â‹\dots 5}{3â‹\dots \left(−7\right)−8â‹\dots 5}\\ =\frac{−305}{−61}\\ =5\end{array}$

$\begin{array}{c}y=\frac{\left|\begin{array}{cc}3& 28\\ 5& −55\end{array}\right|}{\left|\begin{array}{cc}3& 8\\ 5& −7\end{array}\right|}\\ =\frac{3â‹\dots \left(−55\right)−\left(28\right)â‹\dots 5}{3â‹\dots \left(−7\right)−8â‹\dots 5}\\ =\frac{−305}{−61}\\ =5\end{array}$

The obtained value of x is $-4$ and y is 5 so the solution is $\left(-4,5\right)$ that is matching with option B.