Question

Find the value of $k$ for which the given system of equations has infinite number of solutions.

$5x+2y=2k$ and $2(k+1)x+ky=(3k+4)$

• A. 4
• B. 7
• C. 3
• D. 6

Answer 1

The correct option is A 4

Comparing $5x+2y=2k$ with $a_{1}x+b_{1}y+c_1=0$, we get,

$a_1=5,b_1=2,c_1=-2k$

Comparing $2(k+1)x+ky=3k+4$ with $a_{2}x+b_{2}y+c_2=0$, we get,

$a_2=2(k+1),b_2=k,c_2=-(3k+4)$

Now,

Given: the equations has infinite no of solutions.

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

$\frac{5}{2(k+1)}=\frac{2}{k}=\frac{-2k}{-(3k+4)}$

When,

$\frac{5}{2(k+1)}=\frac{2}{k}$

$⟹5k=4(k+1)$

$⟹5k-4k=4$

$⟹k=4$

and

$\frac{2}{k}=\frac{-2k}{-(3k+4)}$

$⟹2(3k+4)=2k^2$

$⟹3k+4=k^2$

$⟹k^2-3k-4=0$

$⟹k^2-4k+k-4=0$

$⟹k(k-4)+(k-4)=0$

$⟹(k-4)(k+1)=0$

Either, $k=4ork=-1$

$k=4$ is the value that satisfies both the conditions.

So, $OptionAiscorrect$