Question

For what value of $k$ does the system of equations $2x+ky=11and5x-7y=5$ has no solution?

• A. $\frac{-14}{5}$
• B. $\frac{-11}{5}$
• C. $\frac{-14}{9}$
• D. $\frac{14}{5}$

Answer 1

The correct option is A $\frac{-14}{5}$

The equations are $2x+ky=11$ and $5x-7y=5$

or $2x+ky-11=0$ and $5x-7y-5=0$

They will have no solution if $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}........\left(1\right)$

Here $a_1=2,a_2=5,b_1=k,b_2=-7,c_1=-11,c_2=-5$

$\therefore \frac{a_1}{a_2}=\frac{2}{5},\frac{b_1}{b_2}=\frac{k}{-7},\frac{c_1}{c_2}=\frac{-11}{-5}=\frac{11}{5}$

Now from $\left(1\right)$

$\frac{a_1}{a_2}=\frac{b_1}{b_2}$

$\therefore \frac{2}{5}=\frac{k}{-7}\Rightarrow k=\frac{-14}{5}$

Also the first two ratios $\ne \frac{c_1}{c_2}$

$\therefore k=\frac{-14}{5}$

Hence, the value of $k$ is $\frac{-14}{5}$