0 karansingh Asked: January 19, 20232023-01-19T00:24:46+05:30 2023-01-19T00:24:46+05:30In: CBSE Find the values of k for which the pair of linear equations kx + 3y = k-2 and 12x + ky = k has no solution. 0 it is important board question cbseclass 10linear equations Share Facebook 1 Answer Voted Oldest Recent [Deleted User] 2023-01-20T10:50:13+05:30Added an answer on January 20, 2023 at 10:50 am The given pair of linear equations is kx + 3y = k – 3 …(i) 12x + ky = k …(ii) On comparing the equations (i) and (ii) with ax + by = c = 0, We get, a_{1} = k, b_{1} = 3, c_{1} = -(k – 3) a_{2} = 12, b_{2} = k, c_{2} = – k Then, a_{1} /a_{2} = k/12 b_{1} /b_{2} = 3/k c_{1} /c_{2} = (k-3)/k For no solution of the pair of linear equations, a_{1}/a_{2} = b_{1}/b_{2}≠ c_{1}/c_{2} k/12 = 3/k ≠ (k-3)/k Taking first two parts, we get k/12 = 3/k k^{2} = 36 k = + 6 Taking last two parts, we get 3/k ≠ (k-3)/k 3k ≠ k(k – 3) k^{2} – 6k ≠ 0 so, k ≠ 0,6 since k is not 6 it must be -6. k=-6 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Add a Video to describe the problem better. Video type Youtube Vimeo Dailymotion Facebook Choose from here the video type. Video ID Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs". Click on image to update the captcha. Save my name, email, and website in this browser for the next time I comment. Related Questions Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back ... . Solve for x and y: 0.4x-1.5y = 6.5, 0.3x-0.2y=0.9. Solve graphically the system of linear equations 4x-5y +16=0 and 2x+y-6 = 0. Determine the vertices of the ... If the remainder on division of x3 + 2x2 + kx +3 by x – 3 is 21, ... Given that x – √5 is a factor of the polynomial x3 – 3√5x2 – 5x + 15√5, ...

The given pair of linear equations is

kx + 3y = k – 3 …(i)

12x + ky = k …(ii)

On comparing the equations (i) and (ii) with ax + by = c = 0,

We get,

a

_{1}= k, b_{1}= 3, c_{1}= -(k – 3)a

_{2}= 12, b_{2}= k, c_{2}= – kThen,

a

_{1}/a_{2}= k/12b

_{1}/b_{2}= 3/kc

_{1}/c_{2}= (k-3)/kFor no solution of the pair of linear equations,

a

_{1}/a_{2}= b_{1}/b_{2}≠ c_{1}/c_{2}k/12 = 3/k ≠ (k-3)/k

Taking first two parts, we get

k/12 = 3/k

k

^{2}= 36k = + 6

Taking last two parts, we get

3/k ≠ (k-3)/k

3k ≠ k(k – 3)

k

^{2}– 6k ≠ 0so, k ≠ 0,6

since k is not 6 it must be -6.

k=-6