91Ó°ÊÓ

The solution of the system of linear equations in two variables:

$\begin{array}{l}{a}_{1}r+b_{1}s=c_1\\ a_{2}r+b_{2}s=c_2\end{array}$

by Cramer’s rule is given by$(r,s)$ where $r=\frac{\left|\begin{array}{cc}{c}_1& b_1\\ c_2& b_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}$$s=\frac{\left|\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}$, and $\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|≠0$.

The given system of linear equations in two variables is:

$\begin{array}{l}7r+5s=3\\ 3r−2s=22\end{array}$

Therefore, by comparing the given system of linear equations with the system of linear .equations $\begin{array}{l}{a}_{1}r+b_{1}s=c_1\\ a_{2}r+b_{2}s=c_2\end{array}$, it can be obtained that:

$a_1=7$, $b_1=5$,$c_1=3$, $a_2=3$, $b_2=-2$ and $c_2=22$.

The value of r is given by:

$\begin{array}{c}r=\frac{\left|\begin{array}{cc}3& 5\\ 22& −2\end{array}\right|}{\left|\begin{array}{cc}7& 5\\ 3& −2\end{array}\right|}\\ =\frac{−6−\left(110\right)}{−14−\left(15\right)}\\ =\frac{−6−110}{−14−15}\\ =\frac{−116}{−29}\\ =4\end{array}$

The value of s is given by:

$\begin{array}{c}s=\frac{\left|\begin{array}{cc}7& 3\\ 3& 22\end{array}\right|}{\left|\begin{array}{cc}7& 5\\ 3& −2\end{array}\right|}\\ =\frac{154−\left(9\right)}{−14−\left(15\right)}\\ =\frac{154−9}{−14−15}\\ =\frac{145}{−29}\\ =−5\end{array}$

The values of x and y are 4 and respectively.

The solution of the given system of equations is $(4,−5)$.