Question

The pair of linear equations $13x+ky=k$ and $39x+6y=k+4$ has infinitely many solutions if:

• A. k = 1
• B. k = 4
• C. k = 6
• D. k = 2

Answer 1

The correct option is D k = 2

The pair of linear equations, $13x+ky-k=0$ and $39x+6y-(k+4)=0$ has infinitely many solutions if:

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

$\Rightarrow \frac{13}{39}=\frac{k}{6}=\frac{-k}{-(k+4)}$

$\Rightarrow \frac{1}{3}=\frac{k}{6}=\frac{k}{k+4}$

Equating the 1st and 2nd part of the above equation we get,

$\frac{1}{3}=\frac{k}{6}$

$\Rightarrow k=2$

OR

Equating the 1st and 3rd part of the above equation we get,

$\frac{1}{3}=\frac{k}{k+4}$

$\Rightarrow k+4=3k$
$\Rightarrow 2k=4$
$\Rightarrow k=2$

$\therefore$ When k = 2, the pair of linear equations has infinite solutions.