Question -
Answer -
(i) 2x + 4y =3
12y + 6x =6
(ii) x =2y
y = 2x
(iii) 3x + y –3 = 0
2x + 2/3y =2
Solution:
The Condition for nosolution = a1/a2 = b1/b2 ≠c1/c2 (parallel lines)
(i) Yes.
Given pair ofequations are,
2x+4y – 3 = 0 and 6x +12y – 6 = 0
Comparing theequations with ax+ by +c = 0;
We get,
a1 =2, b1 = 4, c1 = – 3;
a2 =6, b2 = 12, c2 = – 6;
a1 /a2 =2/6 = 1/3
b1 /b2 =4/12 = 1/3
c1 /c2 =– 3/ – 6 = ½
Here, a1/a2 =b1/b2 ≠ c1/c2, i.eparallel lines
Hence, the given pairof linear equations has no solution.
(ii) No.
Given pair ofequations,
x = 2y or x – 2y = 0
y = 2x or 2x – y = 0;
Comparing theequations with ax+ by +c = 0;
We get,
a1 =1, b1 = – 2, c1 = 0;
a2 =2, b2 = – 1, c2 = 0;
a1 /a2 =½
b1 /b2 =-2/-1 = 2
Here, a1/a2 ≠ b1/b2.
Hence, the given pairof linear equations has unique solution.
(iii) No.
Given pair ofequations,
3x + y – 3 = 0
2x + 2/3 y = 2
Comparing theequations with ax+ by +c = 0;
We get,
a1 =3, b1 = 1, c1 = – 3;
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 coincidentlines