The number of values of `k` for which the system of the equations `(k+1)x+8y=4ka n dk x+(k+3)y=3k-1` has infinitely many solutions is `0` b. `1` c. `2` d. infinite

Options

A. 1
B. 2
C. 3
D. None of the above

Answer

C

Explanation

System of equation is given as :
`(k+1)x+8y = 4k" "...(1)`
`and kx + (k + 3)y = 3k -1" "...(2)`
Here, `a_(1)=k+1, b_(1) = 8, c_(1) = 4k, a_(2) = k, b_(2) = k + 3 and c_(2)=3k - 1`
Such a system of equations will have infinite number of solution, if
`(a_(1))/(a_(2))=(b_(1))/(b_(2))=(c_(1))/(c_(2))`
i.e., `(k+1)/(k)=(8)/(k+3)=(4k)/(3k-1)`
Taking last two we get `8(3k-1)=4k(k+3)`
`rArr 24k - 8 = 4k^(2)+12k`
`rArr 4k^(2)-12k + 8 = 0`
`rArr k^(2)-3k+2=0`
`rArr (k^(2)-3k+2=0`
`rArr (k-1)(k-2)\=0`
`rArr k = 1, 2`
Taking first two `(k+1)(k+3)=8k`
`rArr k^(2)+4k+3-8k=0 rArr k^(2)-4k+3=0`
`rArr (k-1)(k-3)=0`
So, k = 1,3.
Combining both, k = 1, 2, 3.
Thus, this system have 3 values of k.