Question

find the value of k for which kx+2y=k-2;8x+ky=k has infinitely many solutions​

Answer 1

kx + 2y = k - 2

8x + ky = k

The system of equations has no solutions if the lines are parallel, thus:

2y = -kx + k - 2

y = (-k/2)x + (k - 2)/2

ky = -8x + k

y = (-8/k)x + 1

-k/2 = -8/k

k^2 = 16

k = ±4

Also:

(k - 2)/2 ≠ 1

k - 2 ≠ 2

k ≠ 4

k can not be +4 thus the only value for k is -4:

k = -4

Let’s check the results:

y = (-k/2)x + (k - 2)/2

y = 2x + 2 => first equation in slope-intercept form

y = (-8/k)x + 1

y = 2x + 1 => 2nd equation in slope-intercept form

The lines are parallel therefore no solution to the system.

Answer 2

Answer:

The system of equations has no solutions if the lines are parallel, thus:

2y = -kx + k - 2

y = (-k/2)x + (k - 2)/2

ky = -8x + k

y = (-8/k)x + 1

-k/2 = -8/k

k^2 = 16

k = ±4

Also:

(k - 2)/2 ≠ 1

k - 2 ≠ 2

k ≠ 4

k can not be +4 thus the only value for k is -4:

k = -4

Let’s check the results:

can not be +4 thus the only value for k is -4:

k = -4

Let’s check the results:

y = (-k/2)x + (k - 2)/2

y = 2x + 2 => first equation in slope-intercept form

y = (-8/k)x + 1

y = 2x + 1 => 2nd equation in slope-intercept form

The lines are parallel therefore no solution to the system.

The system of equations has no solutions if the lines are parallel, thus:

2y = -kx + k - 2

y = (-k/2)x + (k - 2)/2

ky = -8x + k

y = (-8/k)x + 1

-k/2 = -8/k

k^2 = 16

k = ±4

Also:

(k - 2)/2 ≠ 1

k - 2 ≠ 2

k ≠ 4

k can not be +4 thus the only value for k is -4:

k = -4

Let’s check the results:

can not be +4 thus the only value for k is -4:

k = -4

Let’s check the results:

y = (-k/2)x + (k - 2)/2

y = 2x + 2 => first equation in slope-intercept form

y = (-8/k)x + 1

y = 2x + 1 => 2nd equation in slope-intercept form

The lines are parallel therefore no solution to the system.

The system of equations has no solutions if the lines are parallel, thus:

2y = -kx + k - 2

y = (-k/2)x + (k - 2)/2

ky = -8x + k

y = (-8/k)x + 1

-k/2 = -8/k

k^2 = 16

k = ±4

Also:

(k - 2)/2 ≠ 1

k - 2 ≠ 2

k ≠ 4

k can not be +4 thus the only value for k is -4:

k = -4

Let’s check the results:

k

can not be +4 thus the only value for k is -4:

k = -4

Let’s check the results:

y = (-k/2)x + (k - 2)/2

y = 2x + 2 => first equation in slope-intercept form

y = (-8/k)x + 1

y = 2x + 1 => 2nd equation in slope-intercept form

The lines are parallel therefore no solution to the system.