Mathematics
Grade-8
Easy
Question
Check each of the given systems of equations to see if it has a unique solution, infinitely solutions or no solution.
2x + 3y = 1
3x – y = 7 solutions of the system
- Unique solution
- Infinite solutions
- No solution
- None of the above
Hint:
Check for one of the three conditions
![fraction numerator A 1 over denominator A 2 end fraction equals space fraction numerator B 1 over denominator B 2 end fraction space equals space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space c o i n c i d e n t space
i f space comma space fraction numerator A 1 over denominator A 2 end fraction equals space fraction numerator B 1 over denominator B 2 end fraction space n o t space e q u a l space space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space p a r a l l e l space
e l s e space i f space comma space fraction numerator A 1 over denominator A 2 end fraction space n o t space e q u a l space fraction numerator B 1 over denominator B 2 end fraction space n o t space e q u a l space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space i n t e r s e c t i n g space](https://lh3.googleusercontent.com/nJxangp4sX8CkrXb9hCKqezVkaCRmFh1rzvLfQ7sosUhD9ExMr9spcdRK0pyapc9h5N3O2W1p-bZNXiGKmpa9VyeZNuGC7glhOy1J6gqhSWLx31U2wFDSrv_lZUw0p-fsHwDyt4O1NMcMlYVe9ETfmU)
The correct answer is: Unique solution
Given lines are
2x + 3y = 1
3x – y = 7 solutions of the system
Check for their solution
use relation
![fraction numerator A 1 over denominator A 2 end fraction equals space fraction numerator B 1 over denominator B 2 end fraction space equals space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space c o i n c i d e n t space
i f space comma space fraction numerator A 1 over denominator A 2 end fraction equals space fraction numerator B 1 over denominator B 2 end fraction space n o t space e q u a l space space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space p a r a l l e l space
e l s e space i f space comma space fraction numerator A 1 over denominator A 2 end fraction space n o t space e q u a l space fraction numerator B 1 over denominator B 2 end fraction space n o t space e q u a l space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space i n t e r s e c t i n g space](https://lh3.googleusercontent.com/nJxangp4sX8CkrXb9hCKqezVkaCRmFh1rzvLfQ7sosUhD9ExMr9spcdRK0pyapc9h5N3O2W1p-bZNXiGKmpa9VyeZNuGC7glhOy1J6gqhSWLx31U2wFDSrv_lZUw0p-fsHwDyt4O1NMcMlYVe9ETfmU)
Step 1 :
Checking relation
![2 over 3 space not equal to fraction numerator 3 over denominator negative 1 end fraction not equal to 1 over 7](data:image/png;base64,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)
step 2 :
so , it satisfies the condition of intersection which is 3rd condition
SO , GIVEN lines have exactly one solution because they intersect at one point
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