Question

Determine the value of k so that the following linear equations have no solution:
$(3k+1)x+3y-2=0$
$(k^2+1)x+(k-2)y-5=0$

Answer 1

The given system of equations is
$(3k+1)x+3y-2=0$
$(k^2+1)x+(k-2)y-5=0$

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

where, $a_1=3k+1,b_1=3,c_1=-2$

and $a_1=k^2+1,b_2=k-2,c_2=-5$

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

The given system of equations will have no solution, if

$\frac{3k+1}{k^2+1}=\frac{3}{k-2}\ne \frac{-2}{-5}$

$\Rightarrow \frac{3k+1}{k^2+1}=\frac{3}{k-2}$ and $\frac{3}{k-2}\ne \frac{2}{5}$

Now, $\frac{3k+1}{k^2+1}=\frac{3}{k-2}$

$\Rightarrow (3k+1)(k-2)=3(k^2+1)$

$\Rightarrow 3k^2-5k-2=3k^2+3$

$\Rightarrow -5k-2=3$

$\Rightarrow -5k=5$

$\Rightarrow k=-1$

Clearly, $\frac{3}{k-2}\ne \frac{2}{5}$ for $k=-1$

Hence, the given system of equations will have no solution for $k=-1$.