Show that the following system of equations has a unique solution: $3x + 5y = 12$, $5x + 3y = 4$. Also, find the solution of the given system of equations.

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Show that the following system of equations has a unique solution:
$3x + 5y = 12$ ,
$5x + 3y = 4$ . Also, find the solution of the given system of equations.

Show that the following system of equations has a unique solution:
$3x + 5y = 12$ ,
$5x + 3y = 4$ . Also, find the solution of the given system of equations.

Solution:

The given system of equations is
$3 x+5 y=12$ $5 x+3 y=4$ These equations are of the forms: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+$ where, $a_{1}=3, b_{1}=5$
$a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$
where, $a_{1}=3, b_{1}=5, c_{1}=-12$ and $a_{2}=5, b_{2}=3, c_{2}=-4$
For a unique solution, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , i.e., $\frac{3}{5} \neq \frac{5}{3}$
Therefore, the given system of eq. has a unique solution. Again, the given eq. are:
$\begin{array}{ll} 3 x+5 y=12 \ldots(i) \\ 5 x+3 y=4 \ldots(i) \end{array}$
Multiplying equation(i) by 3 and equation(ii) by 5 , we obtain:
$9 x+15 y=36 \quad \ldots \ldots \text { (iii) }$
$25 x+15 y=20\dots \dots(iv)$
Subtracting equation(iii) from equation(iv), we obtain:
$\begin{array}{l} 16 x=-16 \\ \Rightarrow x=-1 \end{array}$
On substituting $\mathrm{x}=-1$ in equation(i), we obtain:
$\begin{array}{l} 3(-1)+5 y=12 \\ \Rightarrow 5 y=(12+3)=15 \\ \Rightarrow y=3 \end{array}$
As a result, $x=-1$ and $y=3$ is the required solution.