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.