Question

For which value of k will the pair of linear equations have no solution:4x +3y =1 and (k-1)x +4y=3k​

Answer 1

EXPLANATION.

• GIVEN

pair lf linear equation have no solution

4x + 3y = 1 = 4x + 3y - 1 = 0 ......(1)

( k - 1 )x + 4y = 3k = ( k - 1 )x + 4y - 3k = 0 ....(2)

TO FIND VALUE OF K.

conditions for no solution

$\frac{A1}{A2} = \frac{B1}{B2} \ne \frac{C1}{C2}$

A1 = 4, B1 = 3 , c1 = -1

A2 = ( k - 1 ), B2 = 4, c2 = -3k

CASE = 1.

$\frac{a1}{a2} = \frac{b1}{b2}$

$\frac{4}{(k - 1)} = \frac{3}{4}$

$3k \: - 3 = 16$

$3k = 19$

$k = \frac{19}{3}$

CASE = 2.

$\frac{b1}{b2} \ne \frac{c1}{c2}$

$\frac{3}{4} \ne \frac{ - 1}{ - 3k}$

in case 2 it is clearly that not equal

Hence,

The value of k = 19 / 3