 # Are the following pair of linear equations consistent? Justify your 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:

We know that

Conditions for pair of linear equations to be consistent are:

a1/a2 ≠ b1/b2. [unique solution]

a1/a2 = b1/b2 = c1/c2 [coincident or infinitely many solutions]

(i) –3x– 4= 12

4+ 3= 12

No.

The given pair of linear equations

– 3x – 4y – 12 = 0

4y + 3x – 12 = 0

On comparing the given equations with ax + by + c = 0;

We obtain

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/b≠ c1/c2

Hence, the pair of linear equations has no solution, i.e., inconsistent.

(ii) (3/5)x – y = ½

(1/5)x – 3y= 1/6

Yes.

The given pair of linear equations are

(3/5)x – y = ½

(1/5)x – 3y= 1/6

On comparing the given equations with ax + by + c = 0;

We obtain

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 pair of linear equations has unique solution, i.e., consistent.

(iii) 2ax by a

4ax + 2by – 2a = 0; ab ≠ 0

Yes.

The given pair of linear equations are

2ax + by –a = 0

4ax + 2by – 2a = 0

On comparing the given equations with ax + by + c = 0;

We obtain

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 pair of linear equations has infinitely many solutions, i.e., consistent

(iv) x + 3= 11

2 (2+ 6y) = 22

No.

The given pair of linear equations

x + 3y = 11

2x + 6y = 11

On comparing the given equations with ax + by + c = 0;

We obtain

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/b≠ c1/c2.

Hence, the given pair of linear equations has no solution. (0) (0)