Question

The number of values of $k$ for which the system of linear equations,
$(k+2)x+10y=kkx+(k+3)y=k–1$
has no soution, is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$\text{For no solution,}\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}\Rightarrow \frac{k+2}{k}=\frac{10}{k+3}\ne \frac{k}{k-1}\Rightarrow (k+2)(k+3)=10k\Rightarrow k^2-5k+6=0\Rightarrow (k-2)(k-3)=0\Rightarrow k=2,3\text{If}k=2,\text{then both the lines becomes identical. So,}k\ne 2\therefore k=3\text{is the only one value of}k.$