Question -
Answer -
(i) –3x– 4y =12
4y + 3x =12
(ii) (3/5)x – y = ½
(1/5)x – 3y= 1/6
(iii) 2ax + by = a
- ax + 2by – 2a = 0; a, b ≠ 0
(iv) x +3y = 11
2 (2x + 6y)= 22
Solution:
Conditions for pair oflinear equations to be consistent are:
a1/a2 ≠ b1/b2. [uniquesolution]
a1/a2 =b1/b2 = c1/c2 [coincidentor infinitely many solutions]
(i) No.
The given pair of linearequations
– 3x – 4y – 12 = 0 and4y + 3x – 12 = 0
Comparing the aboveequations with ax + by + c = 0;
We get,
a1 = –3, b1 = – 4, c1 = – 12;
a2 =3, b2 = 4, c2 = – 12;
a1 /a2 =– 3/3 = – 1
b1 /b2 =– 4/4 = – 1
c1 /c2 =– 12/ – 12 = 1
Here, a1/a2 =b1/b2 ≠ c1/c2
Hence, the pair oflinear equations has no solution, i.e., inconsistent.
(ii) Yes.
The given pair oflinear equations
(3/5)x – y = ½
(1/5)x – 3y= 1/6
Comparing the aboveequations with ax + by + c = 0;
We get,
a1 =3/5, b1 = – 1, c1 = – ½;
a2 =1/5, b2 = 3, c2 = – 1/6;
a1 /a2 =3
b1 /b2 =– 1/ – 3 = 1/3
c1 /c2 =3
Here, a1/a2 ≠ b1/b2.
Hence, the given pairof linear equations has unique solution, i.e., consistent.
(iii) Yes.
The given pair oflinear equations –
2ax + by –a = 0 and4ax + 2by – 2a = 0
Comparing the aboveequations with ax + by + c = 0;
We get,
a1 =2a, b1 = b, c1 = – a;
a2 =4a, b2 = 2b, c2 = – 2a;
a1 /a2 =½
b1 /b2 =½
c1 /c2 =½
Here, a1/a2 =b1/b2 = c1/c2
Hence, the given pairof linear equations has infinitely many solutions, i.e., consistent
(iv) No.
The given pair oflinear equations
x + 3y = 11 and 2x +6y = 11
Comparing the aboveequations with ax + by + c = 0;
We get,
a1 =1, b1 = 3, c1 = 11
a2 =2, b2 = 6, c2 = 11
a1 /a2 =½
b1 /b2 =½
c1 /c2 =1
Here, a1/a2 =b1/b2 ≠ c1/c2.
Hence, the given pairof linear equations has no solution.