Question

Given $3x-4y=7$ and $x+cy=13$, for what value of $c$ will the two equations not have a solution ?

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $-4$
• D. $\frac{-4}{3}$

Answer 1

The correct option is D $\frac{-4}{3}$

$3x-4y=7$ $\Rightarrow a_1=3,b_1=-4,c_1=7$
$x+cy=13\Rightarrow a_2=1,b_2=c,c_2=13$
As we know the system of linear equation has no solution if

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

$\therefore$ $\frac{3}{1}=\frac{-4}{c}\ne \frac{7}{13}$
$c=\frac{-4}{3}$
Hence, option D is correct.