What is the value of k for which the simultaneous equations 2x 3y 8 and 6x KY 24 have the infinitely many solutions?

The given system of equation is

2x + 3y − 5 = 0

6x + ky − 15 = 0

The system of equation is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where `a_1 = 2, b_1 = 3, c_1 = -5`

And `a_2 = 6, b_2 = k,c_2 = -15`

For a unique solution, we must have

`a_1/a_2 = b_1/b_2 = c_1/c_2`

`=> 2/6 = 3/k`

`=> k = 18/2 = 9`

Hence, the given system of equations will have infinitely many solutions, if k = 9.

Page 2

The given system of equation is

4x + 5y -3 = 0

ks + 15y - 9 = 0

The system of equation is of the for

`a_1x + b_1y + c_1= 0`

`a_2x + b_2y + c_2 = 0`

Where `a_1 = 4, b_1 = 5, c_1 = -3`

And `a_2 = k, b_2 = 15, c_2 = -9`

For a unique solution, we must have

`a_1/a_2 = b_1/b_2 = c_2/c_2`

`=> 4/k = 5/15 = (-3)/(-9)`

Now

`4/k = 5/15`

`=> 4/k = 1/3`

`=> k = 12`

Hence, the given system of equations will have infinitely many solutions, if k = 12

Page 3

The given system of equation is

kx - 2y + 6 = 0

4x + 3y + 9 = 0

The system of equation is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where `a_1 = k, b_1 = -2,c_1 = 6`

And `a_2 = 4, b_2 = -3,c_2 = 9`

For a unique solution, we must have

`a_1/a_2 = b-1/b_2 = c-1/c_2`

`=>k/4 = (-2)/(-3) = 6/9`

Now

`k/4 = 6/9`

`=> k/4 = 2/3`

`=> k = (2xx4)/3`

`=> k = 8/3`

Hence, the given system of equations will have infinitely many solutions, if `k = 8/3`

Page 4

The given system of equation is

8x + 5y - 9 = 0

kx + 10y - 18 = 0

The system of equation is of the form

`a_1x + b_1y + c_1 = 0`

`a_2x + b_2y + c_2 = 0`

Where, `a_1 = 8, b_1 = 5, c_1 = -9`'

And `a_2 = k, b_2 = 10, c_2 = -18`

For a unique solution, we must have

`a_1/a_2 = b_1/b_2 = c_1/c_2`

`=>8/k = 5/10 = (-9)/(-18)`

Now

`8/k = 5/10`

`=> 8 xx 10 = 5 xx k`

`=> (8xx10)/5 = k`

`=> k = 8 xx 2 = 16`

Hence, the given system of equations will have infinitely many solutions, if k = 16

Page 5

The given system of the equation may be written as

2x - 3y - 7 = 0

(k + 2)x - (2k + 1)y - 3(2k -1) = 0

The system of equation is of the form

`a_1x + b_1y + c_1 = 0``

`a_2x + b_2y + c_2 = 0`

Where `a_1 = 2, b_1 = -3, c_1 = -7`

And

`a_2 = k, b_2 = -(2k + 1), c_2 = -3(2k - 1)`

For a unique solution, we must have

`a_1/a_2= b_1/b_2 = c_1/c_2`

`=> 2/(k + 2) = 3/(-(2k + 1)) = (-7)/(-3(2k -1))`

`=> 2/(k+1) = (-3)/(-(2k + 1)) and (-3)/(-(2k + 1)) = (-7)/(-3(2k - 1))` 

`=> 2(2k + 1) =3(k+1)` and `3 xx 3 (2k - 1) = 7(2k + 1)`

`=> 4k + 2 = 3k + 6 and 15k - 9 = 14k + 7`

`=> 4k - 3k = 6 - 2 and 15k - 14k = 7 + 9`

`=> k = 4 and 4k = 16 => k = 4`

`=>k = 4 and k = 4`

Hence, the given system of equations will have infinitely many solutions, if k = 4