Question

For the following system of equation determine the value of k for which the given system has no solution.
$3x-4y+7=0$
$kx+3y-5=0$

Answer 1

The given system of equations is
$3x-4y+7=0$
$kx+3y-5=0$

This is of the form $a_{1}x+b_{1}y+c_1=0$
$a_{2}x+b_{2}y+c_2=0$,

where, $a_1=3,b_1=-4,c_1=7$

and $a_1=k,b_2=3,c_2=-5$

For no solution, we must have
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

We have, $\frac{b_1}{b_2}=\frac{-4}{3}$ and $\frac{c_1}{c_2}=\frac{-7}{5}$

Clearly, $\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

So, the given system will have no solution.

If $\frac{a_1}{a_2}=\frac{b_1}{b_2}\Rightarrow \frac{3}{k}=\frac{-4}{3}\Rightarrow k=\frac{-9}{4}$

Clearly, for this value of k, we have $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Hence, the given system of equations has no solution, when $k=\frac{-9}{4}$.