Question

The value of $k$ for which the equations $9x+4y=9$ and $7x+ky=5$, have no solution is

• A. $\frac{9}{5}$
• B. $\frac{9}{7}$
• C. $\frac{9}{28}$
• D. $\frac{28}{9}$

Answer 1

The correct option is D $\frac{28}{9}$

The given equations are $9x+4y=9$ & $7x+ky=5$.

We know that if there are two equations of the forms

$a_{1}x+b_{1}y+c_1=0$ & $a_{2}x+b_{2}y+c_2=0$

Then, for no solution, $\frac{a_1}{a_2}=\frac{a_1}{a_2}\ne \frac{c_1}{c_2}.$

Here,

$a_1=9,b_1=4,c_1=-9,a_2=7,b_2=k\&c_2=-5\therefore \frac{a_1}{a_2}=\frac{9}{7},\frac{b_1}{b_2}=\frac{4}{k}\&\frac{c_1}{c_2}=\frac{9}{5}.\therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}⟹\frac{9}{7}=\frac{4}{k}⟹k=\frac{28}{9}$