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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

1 Answer

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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/b≠ 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/b≠ c1/c2.

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

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