Question

The system of linear equation $ax+by=6$, $cx+dy=8$ has no solution if:

• A. $ad-bc>0$
• B. $ad-bc<0$
• C. $ad+bc=0$
• D. $ad-bc=0$

Answer 1

Answer: (D)

$ad-bc=0$

The given equations are $ax+by=6$ and $cx+dy=8$.

Here, $a_1=a,b_1=b,c_1=-6and{a}_2=c,b_2=d,c_2=-8$

They have no solution if,

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

Hence, $\frac{a}{c}=\frac{b}{d}$

$\Rightarrow ad-bc=0$

So, option D is correct.