Question

Let the system of linear equations $x+y+ kz =2$ $2 x+3 y-z=1$ $3 x+4 y +2 z= k$ have infinitely many solutions Then the system $( k +1) x+(2 k -1) y=7$ $(2 k +1) x+( k +5) y=10$ has:

• A. infinitely many solutions
• B. unique solution satisfying $x-y=1$
• C. no solution
• D. unique solution satisfying $x+y=1$

Answer 1

Answer: (D)

unique solution satisfying $x+y=1$

Answer 2

∣∣​123​134​k−12​∣∣​=0
⇒1(10)−1(7)+k(−1)−0
⇒k=3
For k=3,2md system is
4x+5y=7....(1)
and 7x+8y=10....(2)
Clearly, they have a unique solution
(2) −(1)⇒3x+3y=3
⇒x+y=1