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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.
x + 2y = 6
2x + 4y = 12 solutions of the system

Unique solution
Infinite solutions
No solution
None of the above

hint Hint:

Check for the relation
$fraction numerator A 1 over denominator A 2 end fraction equals space fraction numerator B 1 over denominator B 2 end fraction space equals space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space c o i n c i d e n t space i f space comma space fraction numerator A 1 over denominator A 2 end fraction equals space fraction numerator B 1 over denominator B 2 end fraction space n o t space e q u a l space space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space p a r a l l e l space e l s e space i f space comma space fraction numerator A 1 over denominator A 2 end fraction space n o t space e q u a l space fraction numerator B 1 over denominator B 2 end fraction space n o t space e q u a l space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space i n t e r s e c t i n g space$

The correct answer is: Infinite solutions

Given lines are
x + 2y = 6
2x + 4y = 12
Check for their solution
use relation
$fraction numerator A 1 over denominator A 2 end fraction equals space fraction numerator B 1 over denominator B 2 end fraction space equals space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space c o i n c i d e n t space i f space comma space fraction numerator A 1 over denominator A 2 end fraction equals space fraction numerator B 1 over denominator B 2 end fraction space n o t space e q u a l space space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space p a r a l l e l space e l s e space i f space comma space fraction numerator A 1 over denominator A 2 end fraction space n o t space e q u a l space fraction numerator B 1 over denominator B 2 end fraction space n o t space e q u a l space fraction numerator C 1 over denominator C 2 end fraction space comma space space t h e n space i t space i s space i n t e r s e c t i n g space$
Step 1 :
Checking relation
$1 half space equals 2 over 4 space equals space 6 over 12$

step 2 :
so , it satisfies the condition of coincident lines which is 1st condition
SO , GIVEN lines have infinity solution because they are same

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