Show that the following system of equations has a unique solution: $2 x-3 y=17$, $4 x+y=13$ Also, find the solution of the given system of equations.

Home » Show that the following system of equations has a unique solution: , Also, find the solution of the given system of equations.

Show that the following system of equations has a unique solution:
$2 x-3 y=17$ , $4 x+y=13$ Also, find the solution of the given system of equations.

Show that the following system of equations has a unique solution:
$2 x-3 y=17$ , $4 x+y=13$ Also, find the solution of the given system of equations.

Solution:

The given system of eq. is:
$\begin{array}{l} 2 x-3 y-17=0\dots \dots(i) \\ 4 x+y-13=\dots \dots(ii) \end{array}$
The given eq. are of the form
$a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$
where, $a_{1}=2, b_{1}=-3, c_{1}=-17$ and $a_{2}=4, b_{2}=1, c_{2}=-13$
So now,
$\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}$ and $\frac{b_{1}}{b_{2}}=\frac{-3}{1}=-3$
As, $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , so the system of eq. has unique solution.
On cross multiplication method, we get:
$\begin{array}{l} \frac{x}{b_{1} c_{2}-b_{2} c_{1}}=\frac{y}{c_{1} a_{2}-c_{2} a_{1}}=\frac{1}{a_{1} b_{2}-a_{2} b_{1}} \\ \Rightarrow \frac{x}{-3(-13)-1 \times(-17)}=\frac{y}{-17 \times 4-(-13) \times 2}=\frac{1}{2 \times 1-4 \times(-3)} \\ \Rightarrow \frac{x}{39+17}=\frac{y}{-68+26}=\frac{1}{2+12} \\ \Rightarrow \frac{x}{56}=\frac{y}{-42}=\frac{1}{14} \\ \Rightarrow x=\frac{56}{14}, y=\frac{-42}{14} \\ \Rightarrow x=4, y=-3 \end{array}$
As a result, $x=4$ and $y=-3$