$$\begin{tabular}{|l|l|} \hline Assertion (A) & Reason $(\mathrm{R})$ \\ \hline The system of equations $\mathrm{x}+\mathrm{y}-8=0$ and $\mathrm{x}-\mathrm{y}-2=0$ has a unique solutions. & $\begin{array}{l}\text { The system of equations } \\ \mathrm{a}_{1} \mathrm{x}+\mathrm{b}_{1} \mathrm{y}+\mathrm{c}_{1}=0 \\ \text { and } \mathrm{a}_{2} \mathrm{x}+\mathrm{b}_{2} \mathrm{y}+\mathrm{c}_{2}-0 \\ \text { has a unique solution when } \\ & \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\end{array}$ \\ \hline \end{tabular}$$ The correct answer is: (a) / (b)/ (c)/ (d).

Home » The correct answer is: (a) / (b)/ (c)/ (d).

$\begin{tabular}{|l|l|} \hline Assertion (A) & Reason $(\mathrm{R})$ \\ \hline The system of equations $\mathrm{x}+\mathrm{y}-8=0$ and $\mathrm{x}-\mathrm{y}-2=0$ has a unique solutions. & $\begin{array}{l}\text { The system of equations } \\ \mathrm{a}_{1} \mathrm{x}+\mathrm{b}_{1} \mathrm{y}+\mathrm{c}_{1}=0 \\ \text { and } \mathrm{a}_{2} \mathrm{x}+\mathrm{b}_{2} \mathrm{y}+\mathrm{c}_{2}-0 \\ \text { has a unique solution when } \\ & \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\end{array}$ \\ \hline \end{tabular}$

The correct answer is: (a) / (b)/ (c)/ (d).

$\begin{tabular}{|l|l|} \hline Assertion (A) & Reason $(\mathrm{R})$ \\ \hline The system of equations $\mathrm{x}+\mathrm{y}-8=0$ and $\mathrm{x}-\mathrm{y}-2=0$ has a unique solutions. & $\begin{array}{l}\text { The system of equations } \\ \mathrm{a}_{1} \mathrm{x}+\mathrm{b}_{1} \mathrm{y}+\mathrm{c}_{1}=0 \\ \text { and } \mathrm{a}_{2} \mathrm{x}+\mathrm{b}_{2} \mathrm{y}+\mathrm{c}_{2}-0 \\ \text { has a unique solution when } \\ & \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\end{array}$ \\ \hline \end{tabular}$

The correct answer is: (a) / (b)/ (c)/ (d).

Answer: (c)

Solution:
The correct answer is option (C).
It is clear that, Reason (R) is false.
Upon solving $x+y=8$ and $x-y=2$ , we obtain:
$x=5$ and $y=3$
Therefore, the given system has a unique solution. So, assertion (A) is true.
$\therefore$ Assertion (A) is true and Reason (R) is false.