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Is it true that the equations $- 3 x– 4 y = 12$ and $4 y + 3 x = 12$ have a unique solution?

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True

False

answer is B.

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Detailed Solution

Given equations are

- 3 x– 4 y = 12

and

4 y + 3 x = 12

.
We know that a pair of linear equations where

a, b ≠ 0,

has a unique solution if,

a 1 a 2 ≠ b 1 b 2

.
And, has no solution if,

a 1 a 2 = b 1 b 2 ≠ c 1 c 2

.
From the linear equation

− 3 x − 4 y = 12 and 3 x + 4 y = 12,

we get the values as,

a 1 = − 3, b 1 = − 4, c 1 = 12 and a 2 = 3, b 2 = 4, c 2 = 12.

The ratios of the coefficients are,

a 1 a 2 = − 3 3 ⇒ a 1 a 2 = − 1, b 1 b 2 = − 4 4 ⇒ b 1 b 2 = − 1, c 1 c 2 = 1212 ⇒ c 1 c 2 = 1

Since, these values satisfy the relation

a 1 a 2 = b 1 b 2 ≠ c 1 c 2 .

So, the given statement is false.
The correct statement is, the pair of linear equations

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

is inconsistent and has no solutions.
Hence, the correct option is 2.

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