| Assertion (A): The system of equations \[x+y-6=0\] and \[\text{x}-\text{y}-\text{2}=0\]has a unique solution. |
| Reason (R): The system of equations \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\]and \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\] has a unique solution when \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\]. |
A) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
B) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
C) Assertion (A) is true but reason (R) is false.
D) Assertion (A) is false but reason (R) is true.
Correct Answer: C
Solution :
| [c] The given system of equations is |
| \[x+y-6=0\] and \[x-y-2=0\] |
| Here, \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{1}=1,\] \[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{-1}=-1,\]\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-6}{-2}=3\] |
| Since, \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\] |
| \[\therefore \] The given system of equations has a unique solution. |
| So, Assertion: True; Reason: False. |
You need to login to perform this action.
You will be redirected in
3 sec
