Question

Find the value of $k$ for which the given simultaneous equations have infinitely many solutions:

$kx-y+3-k=0;4x-ky+k=0$

• A. $k=3$
• B. $k=4$
• C. $k=2$
• D. $k=1$

Answer 1

The correct option is C $k=2$

Compare the equations $kx-y+3-k=0$ and $4x-ky+k=0$ with the general equations $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ respectively.

Here, $\frac{a_1}{a_2}=\frac{k}{4},\frac{b_1}{b_2}=\frac{-1}{-k}=\frac{1}{k}$ and $\frac{c_1}{c_2}=\frac{3-k}{k}$.

For a pair of linear equations to have infinitely many solutions : $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So, we have $\frac{k}{4}=\frac{1}{k}=\frac{3-k}{k}$

We first solve $\frac{k}{4}=\frac{1}{k}$

which gives $k^2=4$, i.e., $k=\pm 2$.

Also, $\frac{1}{k}=\frac{3-k}{k}$

gives $k=3k-k^2$, i.e., $k^2-2k=0$, which means $k=0$ or $k=2$.

Therefore, the value of $k$, that satisfies both the conditions, is $k=2$.

Hence,for $k=2$, the pair of linear equations has infinitely many solutions.