Question

Find x and y using Cramer's Rule,if $\frac{1}{x}-\frac{2}{y}=6,\frac{3}{x}+\frac{1}{y}=8$ ?

Answer 1

Given the two equations, $\frac{1}{x}-\frac{2}{y}=6⟶\left(1\right)$

$\frac{3}{x}+\frac{1}{y}=8⟶\left(2\right)$

Let, $p=\frac{1}{x},q=\frac{1}{y}$.

Then, equation $\left(1\right)$ becomes$\Rightarrow p-2q=6⟶\left(3\right)$

and equation $\left(2\right)$ becomes$\Rightarrow 3p+q=8⟶\left(4\right)$.

Now, solving equation $\left(3\right)$ and $\left(4\right)$ by Cramer's rule,

Let, $A=\left(\begin{array}{cc}1& -2\\ 3& 1\end{array}\right)$

Finding the determinant of the matrix $A$,

$\left|A\right|=\left|\begin{array}{cc}1& -2\\ 3& 1\end{array}\right|$

$=1+6=7$

Now,for matrix $A_p$, we have to replace the first column of the matrix $A$ with the right hand side values of equation $\left(3\right)$ and $\left(4\right)$, $A_p=\left(\begin{array}{cc}6& -2\\ 8& 1\end{array}\right)$

$\therefore \left|A_p\right|=\left|\begin{array}{cc}6& -2\\ 8& 1\end{array}\right|$

$=6+16=22$

$\therefore p=\frac{\left|A_p\right|}{\left|A\right|}=\frac{22}{7}$

Again, for matrix $A_q$, we have to replace the second column of the matrix $A$ with the right hand side values of equation $\left(3\right)$ and $\left(4\right)$,$A_q=\left(\begin{array}{cc}1& 6\\ 3& 8\end{array}\right)$

$\therefore \left|A_q\right|=\left|\begin{array}{cc}1& 6\\ 3& 8\end{array}\right|=8-18=-10$

$\therefore q=\frac{\left|A_q\right|}{\left|A\right|}=\frac{-10}{7}$

Thus, $\frac{1}{x}=\frac{22}{7},\frac{1}{y}=\frac{-10}{7}$

$\Rightarrow x=\frac{7}{22},y=-\frac{7}{10}$

Hence, the solutions are $x=\frac{7}{22}andy=-\frac{7}{10}$.