Question

for which value of k will the pair of linear equations kx+3y=k-3 and 12x+ky=k have no solution?

Answer 1

thats alot of ks ...

Answer 2

in effect; when 2 linear equations like that have the same left side, but different right sides; they never are equal

Answer 3

For no Solution these two lines(equations) shud become parallel to each other,multiply their slopes and equate that to 1
m1*m2=1

Answer 4

And find k

Answer 5

kx + 3y = 12x + ky

kx + 3y
-12x - ky
---------------
(k-12)x + (3-k)y = 0


umm, if they have the same slope, them m1*m2 will not equate to 1 in general will it?

Answer 6

oh,i forgot the condition for parallel lines.m1=m2

Answer 7

right? @amistre64

Answer 8

em nt ggttng d answer guyzzzzz

Answer 9

yes :) or to use your vernacular \(m_1*m_2^{-1}=1\) :)

Answer 10

what have you tried?

Answer 11

kx + 3y
-12x - ky
---------------
(k-12)x + (3-k)y = 0

Answer 12

after dt???????////

Answer 13

cn u plzzz show it after dt i will telll where i went wrong

Answer 14

just an idea; but another might to be to determine when they are both equal to each other

kx+3y-k+3 = 12x+ky -k

kx-k = 12x -3y -3 +ky -k

kx-k -ky +k = 12x -3y -3

k(x -1 -y +1) = 12x -3y -3

\[ k=\frac{12x -3y -3}{x-y}\]

Answer 15

m1=-k/3, m2=-12/k Find thier slopes by differentiating(its easy)
-k/3=-12/k
k^2=36
k=+ or - 6

Answer 16

we are allowed to use calculus? hmmm.....

Answer 17

im thinking matrix linear algebra stuff in the end :)

Answer 18

thx guyzzzzzzzzzz

Answer 19

first re-arrange both equation in slope-intercept form to get:\[y=-\frac{kx}{3}+\frac{k-3}{3}\tag{1}\]and:\[y=-\frac{12x}{k}+1\tag{2}\]now, for no solutions, both lines must have the same slope but different y-intercepts. If they had the same y-intercept then there are an infinite number of solutions.
so, with these conditions we get:\[-\frac{k}{3}=-\frac{12}{k}\implies k^2=36\implies k=\pm6\tag{3}\]and:\[\frac{k-3}{3}\ne1\implies k-3\ne3\implies k\ne 6\tag{4}\]now combine the results of (3) and (4) to find the answer.