Question

3. 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
4 ax + 2by – 2a = 0; a, b ≠ 0
(iv) x + 3y = 11
2 (2x + 6y) = 22

Answer 1

Answer:

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) No.

The given pair of linear equations

- 3x - 4y - 12 = 0 and 4y + 3x - 12 = 0

Comparing the above equations 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 of linear equations has no solution, i.e., inconsistent.

(ii) Yes.

The given pair of linear equations

(3/5)x – y = ½

(1/5)x – 3y= 1/6

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

(iii) Yes.

The given pair of linear equations –

2ax + by –a = 0 and 4ax + 2by - 2a = 0

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

(iv) No.

The given pair of linear equations

x + 3y = 11 and 2x + 6y = 11

Comparing the above equations 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 pair of linear equations has no solution