Question

Find the value of $k$ for which the system of linear equation $kx+4y=k-4$ and $16x+ky=k$ has infinite solution.

• A. $k=7$
• B. $k=13$
• C. $k=8$
• D. $k=15$

Answer 1

Answer: (C)

$k=8$

Given equations are $kx+4y=k-4$ and $16x+ky=k$

Thus, we have

$a_1=k,b_1=4,c_1=-(k-4)$
$a_2=16,b_2=k,c_2=-k$
Here condition is $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\Rightarrow$ $\frac{k}{16}=\frac{4}{k}=\frac{(k-4)}{\left(k\right)}$
$\Rightarrow$ $\frac{k}{16}=\frac{4}{k}$
Also $\frac{4}{k}=\frac{k-4}{k}$
$\Rightarrow$ $k^2=64$
$\Rightarrow$ $4k=k^2-4k$
$\Rightarrow$ $k=\pm 8$
$\Rightarrow$ $k^2-8k=0$
$\Rightarrow$ $k(k-8)=0$
$k=0$ or $k=8$ but $k=0$ is not possible other wise equation will be one variable.
$\Rightarrow$ $k=8$ is correct value for infinite solution.