Question

for what value of'k; the pair of linear equations KX+3y=1 and 12+ky=2 has no solution.​

Answer 1

Answer:

STEP BY STEP EXPLANATION

kx + 3y - 1= 0 and 12x + ky - 2 =0

Here,

a1 = k1 , b1 = 3 , c1 = -1

a2 = 12, b2 = k , c2= -2

The equation for no solution:

$\frac{a1}{a2}$

=

$\frac{b1 }{b2}$

is not equal to

$\frac{c1}{c2}$

Now , substituting the values

$\frac{k}{12}$

=

$\frac{3}{k}$

${k}^{2}$

= 36

Therefore K= +6 or K= -6

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