Are the following pair of linear equations consistent justify your answer 3x -- 4y = 12 4y + 3x = 12

Conditions for pair of linear equations to be consistent are:

`a_1/a_2 ≠ b_1/b_2`.  ......[Unique solution]

`a_1/a_2 = b_1/b_2 = c_1/c_2`   ......[Coincident or infinitely many solutions]

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

`a_1/a_2 = - 3/3 = -1`

`b_1/b_2 = - 4/4 = -1`

`c_1 /c_2 = (-12)/-12` = 1

Here, `a_1/a_2 = b_1/b_2 ≠ c_1/c_2`

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

Page 2

Conditions for pair of linear equations to be consistent are:

`a_1/a_2 ≠ b_1/b_2` ......[Unique solution]

`a_1/a_2 = b_1/b_2 = c_1/c_2`......[Coincident or infinitely many solutions]

Yes.

The given pair of linear equations

`(3/5)x - y = 1/2`

`(1/5)x - 3y = 1/6`

Comparing the above equations with ax + by + c = 0;

We get,

`a_1 = 3/5, b_1 = -1, c_1 = -1/2`

`a_2 = 1/5, b_2 = 3, c_2 = -1/6`

`a_1/a_2` = 3

`b_1/b_2 = (-1)/-3 = 1/3`

`c_1/c_2` = 3

Here, `a_1/a_2 ≠ b_1/b_2`.

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

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Conditions for pair of linear equations to be consistent are:

`a_1/a_2 ≠ b_1/b_2` ......[Unique solution]

`a_1/a_2 = b_1/b_2 = c_1/c_2`......[Coincident or infinitely many solutions]

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

`a_1/a_2 = 1/2`

`b_1/b_2 = 1/2`

`c_1/c_2 = 1/2`

Here, `a_1/a_2 = b_1/b_2 = c_1/c_2`

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

Page 4

Are the following pair of linear equations consistent? Justify your answer.

x + 3y = 11, 2(2x + 6y) = 22

Conditions for pair of linear equations to be consistent are:

`a_1/a_2 ≠ b_1/b_2` ......[Unique solution]

`a_1/a_2 = b_1/b_2 = c_1/c_2`  ......[Coincident or infinitely many solutions]

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

`a_1/a_2 = 1/2`

`b_1/b_2 = 1/2`

`c_1/c_2` = 1

Here, `a_1/a_2 = b_1/b_2 ≠ c_1/c_2`.

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

Concept: Consistency of Pair of Linear Equations

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