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 areeb123xyz 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 â€“ 3 â€¦(i) 12x + ky = k â€¦(ii) On comparing the equations (i) and (ii) with ax + by = c = 0, We get, a_{1}Â =Â k, b_{1}Â = 3, c_{1}Â = -(k â€“ 3) a_{2}Â = 12, b_{2}Â =Â k, c_{2}Â = â€“ k Then, a_{1}Â /a_{2}Â =Â k/12 b_{1}Â /b_{2}Â = 3/k c_{1}Â /c_{2}Â = (k-3)/k For no solution of the pair of linear equations, a_{1}/a_{2}Â = b_{1}/b_{2}â‰ Â c_{1}/c_{2} k/12Â = 3/k â‰ (k-3)/k Taking first two parts, we get k/12 = 3/k k^{2}Â = 36 k = +Â 6 Taking last two parts, we get 3/k â‰ (k-3)/k 3kÂ â‰ Â k(k â€“ 3) k^{2}Â â€“ 6kÂ â‰ 0 so, kÂ â‰ Â 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 2. Write the arithmetic progression when the first term a and common difference d are as follows: (i) ... 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 â€“ 3 is 21, ...

The given pair of linear equations is

kx + 3y = k â€“ 3 â€¦(i)

12x + ky = k â€¦(ii)

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

We get,

a

_{1}Â =Â k, b_{1}Â = 3, c_{1}Â = -(k â€“ 3)a

_{2}Â = 12, b_{2}Â =Â k, c_{2}Â = â€“ kThen,

a

_{1}Â /a_{2}Â =Â k/12b

_{1}Â /b_{2}Â = 3/kc

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

a

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

Taking first two parts, we get

k/12 = 3/k

k

^{2}Â = 36k = +Â 6

Taking last two parts, we get

3/k â‰ (k-3)/k

3kÂ â‰ Â k(k â€“ 3)

k

^{2}Â â€“ 6kÂ â‰ 0so, kÂ â‰ Â 0,6

since k is not 6 it must be -6.

k=-6