Question

If the pair of linear equations $3x+5y=3$ and $6x+ky=8$ do not have any solution, which of the following is true?

• A. $k=5$
• B. $k=10$
• C. $k\ne 10$
• D. $k\ne 5$

Answer 1

Answer: (B)

$k=10$

$3x+5y=3$

$6x+ky=8$

Comparing $3x+5y=3$ with $a_{1}x+b_{1}y+c_1=0$, we get, $a_1=3,b_1=5,c_1=-3$

Comparing$6x+ky=8$ with $a_{2}x+b_{2}y+c_2=0$, we get,$a_2=6,b_2=k,c_2=-8$

For no solutions , we know,

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

$\therefore \frac{3}{6}=\frac{5}{k}\ne \frac{3}{8}$

$\therefore \frac{1}{2}=\frac{5}{k}$

$⟹k=10$ is true