Question

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

• A. $k=-1$
• B. $k=3$
• C. $k=-7$
• D. $k=8$

Answer 1

The correct option is A $k=-1$

Given equations are $(3k+1)x+3y-2=0$ and $(k^2+1)x+(k-2)y-5=0$

Here $a_1=3k+1,b_1=3$ and $c_1=-2$
$a_2=k^2+1,b_2=k-2$ and $c_2=-5$
For no solution, condition is $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{3k+1}{k^2+1}=\frac{3}{k-2}$
$\Rightarrow$ $\frac{3k+1}{k^2+1}=\frac{3}{k-2}$
$\Rightarrow (3k+1)(k-2)=3(k^2+1)$
$\Rightarrow$ $3k^2-5k-2=3k^2+3$
$\Rightarrow$ $-5k-2=3$
$\Rightarrow -5k=5$
$\Rightarrow k=-1$
Clearly, $\frac{3}{k-2}\ne \frac{2}{5}$ for $k=-1$
Hence, the given system of equations will have no solution for $k=-1$.