Do the following pair of linear equations have no solution? Justify your answer.

(i) 2x + 4y = 3 and 12y + 6x = 6


(ii) x = 2y and y = 2x


(iii) 3x + y - 3 = 0 and


The Condition for no solution is : (parallel lines)

(i) Yes.


Given pair of equations are,


2x+4y - 3 = 0 and 6x + 12y - 6 = 0


Comparing with ax+ by +c = 0;


Here, a1 = 2, b1 = 4, c1 = - 3;


And a2 = 6, b2 = 12, c2 = - 6;


a1 /a2 = 2/6 = 1/3


b1 /b2 = 4/12 = 1/3


c1 /c2 = - 3/ - 6 = 1/2


Here, a1/a2 = b1/b2 c1/c2, i.e parallel lines


Hence, the given pair of linear equations has no solution.


(ii) No.


Given pair of equations,


x = 2y and y = 2x


or x - 2y = 0 and 2x - y = 0;


Comparing with ax + by + c = 0;


Here, a1 = 1, b1 = - 2, c1 = 0;


And a2 = 2, b2 = - 1, c2 = 0;


a1 /a2 = 1/2


b1 /b2 = -2/-1 = 2


Here, a1/a2 b1/b2.


Hence, the given pair of linear equations has unique solution.


(iii) No.


Given pair of equations,


3x + y - 3 = 0


and


Comparing with ax + by + c = 0;


Here, a1 = 3, b1 = 1, c1 = - 3;


And a2 = 2, b2 = 2/3, c2 = - 2;


a1 /a2 = 2/6 = 3/2


b1 /b2 = 4/12 = 3/2


c1 /c2 = - 3/-2 = 3/2


Here, a1/a2 = b1/b2 = c1/c2, i.e coincident lines


Hence, the given pair of linear equations is coincident and having infinitely many solutions.


1
1