Question

The number of values of $k$ for which the system of equations $(k+1)x+8y=4k;kx+(k+3)y=3k–1$ has infinitely many solutions is

• A. $0$
• B. $1$
• C. $2$
• D. Infinite

Answer 1

The correct option is B

$1$

The explanation for the correct answer.

Solve for the value of $k$

For infinitely many solutions, the ratio of coefficient will be same.

$$\begin{gathered} \frac{k+1}{k}=\frac{8}{k+3}=\frac{4}{3k-1} \\ \Rightarrow \frac{8}{k+3}=\frac{4}{3k-1} \\ \Rightarrow 24k-8=4k+12 \\ \Rightarrow 20k=20 \\ \Rightarrow k=1 \end{gathered}$$

Hence, option(B) is the correct answer.