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, a_{1} = 2, b_{1} = 4, c_{1} = - 3;

And a_{2} = 6, b_{2} = 12, c_{2} = - 6;

a_{1} /a_{2} = 2/6 = 1/3

b_{1} /b_{2} = 4/12 = 1/3

c_{1} /c_{2} = - 3/ - 6 = 1/2

Here, a_{1}/a_{2} = b_{1}/b_{2} c_{1}/c_{2}, 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, a_{1} = 1, b_{1} = - 2, c_{1} = 0;

And a_{2} = 2, b_{2} = - 1, c_{2} = 0;

a_{1} /a_{2} = 1/2

b_{1} /b_{2} = -2/-1 = 2

Here, a_{1}/a_{2} b_{1}/b_{2.}

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, a_{1} = 3, b_{1} = 1, c_{1} = - 3;

And a_{2} = 2, b_{2} = 2/3, c_{2} = - 2;

a_{1} /a_{2} = 2/6 = 3/2

b_{1} /b_{2} = 4/12 = 3/2

c_{1} /c_{2} = - 3/-2 = 3/2

Here, a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}, i.e coincident lines

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

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