Here, the equations are; (k+2)x+10y=k &kx+(k+3)y=k−1. These equations can be written in the form of Ax=B as [
k+2
10
k
k+3
][
x
y
]=[
k
k−1
] For the system to have no solution |A|=0 ⇒[
k+2
10
k
k+3
]=0⇒(k+2)(k+3)−k×10=0 ⇒k2−5k+6=(k−2)(k−3)=0 ∴k=2,3 For k=2, equations become: 4x+10y=2 &2x+5y=1 & hence infinite number of solutions. For k=3, equations becomes; 5x+10y=3 3x+6y=2 & hence no solution. ∴ required number of values of k is 1