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