Question

find the value of k for which kx+2y=k-2 and 8x+ky=16 has no solution question no.8 in pic

Answer 1

Please see the attachment

Answer 2

As per the data given in the given Question.

We have to find the value of k and given equation have no solution.

Given,

$kx+2y=k-2$

$8x+ky=16$

Step-by-step explanation:

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

$\frac{a_1}{a_2} = \frac{b_1}{b_2} ≠ \frac{c_1}{c_2}$

Given Equation (1)

$kx+2y=k-2 \: \: \: \: \: ...(1)$

shift the value ,then

$2y = -kx + k - 2$

Now , shift the 2 then becomes

$y = ( \frac{k}{2}) x + \frac{ (k - 2)}{2} \: \: \: \: ......(2)$

Given Equation (2)

$8x+ky=16 \: \: \: \: \: \: ....(3)$

Shift the terms to get the value of y

$ky = -8x + k$

$y = ( \frac{ - 8}{k} )x + 1 \: \: \: \: \: ...(4)$

Now , compare the first term of equation (2) and (4)

$\frac{ - k}{2} = \frac{ - 8}{k}$

$k^2 = 16$

$k = ±4$

Also compare the second term of equation (2) and (4)

$\frac{k - 2}{2} ≠ 1$

$k - 2 ≠ 2$

$k ≠ 4$

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

$k = -4$

Hence ,

The lines are parallel therefore no solution to the system and value is +4 .

Project code #SPJ3