Question

If the system $2x+3y-5=0$, $4x+ky-10=0$ has an infinite number of solutions then:

• A. $k=\frac{3}{2}$
• B. $k\ne \frac{3}{2}$
• C. $k\ne 6$
• D. $k=6$

Answer 1

The correct option is D $k=6$

Given system of linear equations:

$2x+3y-5=0$
$4x+ky-10=0$

The above system has infinite number of solutions, as given in the question.

Comparing the given system of equations with $a_{1}x+b_{1}y+c_1=0\text{and}{a}_{2}x+b_{2}y+c_2=0$, we get:

$a_1=2,b_1=3,c_1=-5$
$a_2=4,b_2=k,c_2=-10$

We know that, for a system of two linear equations to have infinite solution:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\therefore \frac{2}{4}=\frac{3}{k}=\frac{-5}{-10}$

$\Rightarrow \frac{1}{2}=\frac{3}{k}=\frac{1}{2}$

Taking
$\frac{3}{k}=\frac{1}{2}$
$\Rightarrow k=6$

Hence, if the given system of linear equations has infinite number of solutions, then $k=6$.