Find k if the following system of equations has no solution 3x+2y=6 and kx+(k-1)y=9?

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Find k if the following system of equations have infinite solutions kx+3y=k-3 and 12x+ky=k

Answer 1

kx + (k - 1)y = 9
(k - 1)y = -kx + 9
y = (-k / k-1)x + 9/(k - 1)

3x + 2y = 6
2y = -3x + 6
y = (-3/2)x + 3

If there is no solution, the two lines must be parallel - and therefore have the same slope. Since I put both in slope intercept form, you just set the slopes equal.

(-k / k-1) = -3/2
(-k)(2) = (-3)(k -1)
-2k = -3k + 3
k = 3

Answer 2

i can see by technique of inspection that ok=3 will do, yet enable's justify the answer. If the gadget of equations has no answer, it represents parallel lines. Parallel lines have a similar slope, yet diverse y-intercepts. Write both equations in y=mx+b style to discover their slopes: 3x+2y=6 2y=6-3x y=(-3/2)x+3 kx+(ok-a million)y=9 (ok-a million)y=-kx+9 y=(-ok/(ok-a million))x+9/(ok-a million) Make both slopes equivalent to at least one yet another: -3/2 = -ok/(ok-a million) 3(ok-a million) = 2k 3k-3 = 2k ok = 3 examine contained in the unique equation. placing ok=3 provides a 2d equation 3x+2y=9, it really is clearly parallel to 3x+2y=6.

Answer 3

A long way :
2y=6-x
y=3-x/2

kx+(k-1)(3-x/2)=9
kx+3k-kx/2-3+x/2=9
kx-kx/2=9+3-3k
kx/2=12-3k
x=(12-3k)2/k
x=(24-6k)/k

for k=0 no x value.

Answer 4

for no solution the condition is 3/k equal to 2/(k-1) not equal to 6/9
solving 3(k-1) not equal to 2k
or k not equal to 3

Answer 5

k = 3

they have the same coefficient of variables but different constants...

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