Question

Find the value of $k$ for which the following system of linear equations has infinite solutions:

$$\begin{gathered} x+(k+1)y=5 \\ (k+1)x+9y=8k-1 \end{gathered}$$

• A. $3$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C

$2$

Explanation for the correct option

Given system of equations is

$\begin{array}{cc}x+(k+1)y=5& \left(\mathrm{i}\right)\\ (k+1)x+9y=8k-1& \left(\mathrm{ii}\right)\end{array}$

Comparing with general form $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$

We get $a_1=1,b_1=k+1,c_1=5$

$a_2=k+1,b_2=9,c_2=8k-1$

We know that, for a pair of linear equations to have infinitely many solutions,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\Rightarrow \frac{1}{(k+1)}=\frac{(k+1)}{9}=\frac{5}{(8k-1)}\left(iii\right)$

Take first term and second term in $\left(iii\right)$

$\begin{array}{ccc}& \frac{1}{(k+1)}=\frac{(k+1)}{9}& \\ \Rightarrow & {(k+1)}^2=9& \\ \Rightarrow & k+1=3,& k+1=-3\\ \Rightarrow & k=2& k=-4\end{array}$

Take first and last $\left(iii\right)$

$\begin{array}{cc}& \frac{1}{(k+1)}=\frac{5}{(8k-1)}\\ \Rightarrow & 5(k+1)=8k-1\\ \Rightarrow & 5k+5=8k-1\\ \Rightarrow & 3k=6\\ \Rightarrow & k=2\end{array}$

So, when $k=2$ the given system of equations has infinite solutions.

Hence, option C is correct.