Question

please answer this question ​

Answer 1

The given equations are,

3x + y = 1

(2k-1)x + (k-1)y = 2k + 1

The given pair of equation will be inconsistent if,

3/2k-1 = 1/k-1

=> 3(k-1) = 2k - 1

=> 3k - 3 = 2k - 1

=> 3k-2k = 3 - 1

=> K = 2

Hence, for K = 2 ,the given equation will be inconsistent.

hope it helps.. ...

Answer 2

Question:

The value of k for which the system of equations 3x + y = 1 and (2k - 1)x + (k - 1)y = 2k + 1 is inconsistent, is?

Answer:

Given equations,

3x + y = 1 ...... (i)

(2k - 1)x + (k - 1)y = 2k + 1 ....... (ii)

Comparing the given equations with $a_1x + b_1y = c_1\:\&\: a_2x + b_2y = c_2$ we get,

a₁ = 3, b₁ = 1, c₁ = 1, a₂ = (2k - 1), b₂ = (k - 1), c₂ = 2k + 1

The conditions for getting inconsistent values or parallel lines is,

$\large{\bold{\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}}}$

Putting up the values,

$\frac{3}{(2k - 1)} = \frac{1}{(k - 1)} \neq \frac{1}{2k + 1}$

Comparing $\frac{3}{(2k - 1)} \:\&\: \frac{1}{(k - 1)}$

$\frac{3}{(2k - 1)} = \frac{1}{(k - 1)}$

3(k - 1) = 1(2k - 1) (cross - multiplying)

3k - 3 = 2k - 1

3k - 2k = - 1 + 3

$\boxed{\bold{\mathsf{k = 2}}}$

We can't compare $\frac{b_1}{b_2}\:and\:\frac{c_1}{c_2}$ as they aren't equal.

So, the value of k is 2.