Math, asked by princy7671, 9 months ago

find the value of kfor which each of the following pair of linear equation has no solution 2x-ky+3=0 , 3x+2y-1=0

Answers

Answered by smtirth

Answer:

a1/a2=b1/b2

If this will be there then the linear equation have no solution.

Therefore, 2x-ky+3=0, 3x+2y-1=0

Now, a1=2 and a2=3

And, b1=-k and b2=2

Then k=3

Because if b1=-3 and b2=2,

-3/2=2/3

So k=3 and -k=-3

Hope my ans. will help u

Pls mark me as brainliest

Previous Question

Next Question

Similar questions

Social Sciences, 4 months ago

Math, 4 months ago

Math, 4 months ago

Hindi, 9 months ago

English, 9 months ago

Chemistry, 11 months ago

Chemistry, 11 months ago

In which of the following reactions heterogeneous catalysis is involved? (i) $2SO_2(g) + O_2(g) \xrightarrow[]{NO(g)} 2SO_3(g)$ (ii) $2SO_2(g) \xrightarrow[]{Pt(s)} 2SO_3(g)$ (iii) $N_2(g) + 3 H_2(g) \xrightarrow[]{Fe(s)} 2NH_3(g)$ (iv) $CH_3COOCH_3(l) + H_2O(l) \xrightarrow[]{HCl(l)} CH_3COOH(aq) + CH_3OH(aq)$ (a) (ii), (iii) (b) (ii), (iii) and (iv)...

Chemistry, 11 months ago