For what value of k will …

Given linear equation is
(3k + 1)x + 3 y - 2 = 0 .......... (i)
(k² + 1)x + (k - 2)y - 5 = 0 ............ (ii)
Compare with a₁x + b₁y + c = 0 and a₂x + b₂y and c₂ = 0
a₁ = 3k + 1 , b₁ = 3 , c₁ = -2
and a₂ = k²+ 1 , b₂ = k - 2, c₂ = -5
The given system of equations will have no solution, if
{tex} \frac { a_1 } { a_2 } = \frac { b_1 } { b_2} \neq \frac { c_1 } { c_2 }{/tex}
{tex} \frac { 3 k + 1 } { k ^ { 2 } + 1 } = \frac { 3 } { k - 2 } \neq \frac { - 2 } { - 5 }{/tex}
{tex}\Rightarrow \quad \frac { 3 k + 1 } { k ^ { 2 } + 1 } = \frac { 3 } { k - 2 } \text { and } \frac { 3 } { k - 2 } \neq \frac { 2 } { 5 }{/tex}
Now, {tex}\frac { 3 k + 1 } { k ^ { 2 } + 1 } = \frac { 3 } { k - 2 }{/tex}
{tex}\Rightarrow{/tex} (3k + 1)(k - 2)=3(k² + 1)
{tex}\Rightarrow{/tex} 3k² - 5k - 2 =3k² + 3
{tex}\Rightarrow{/tex} -5k - 2 =3
{tex}\Rightarrow{/tex} -5k = 5
{tex}\Rightarrow{/tex} k = -1
Clearly, {tex}\frac { 3 } { k - 2 } \neq \frac { 2 } { 5 }{/tex} for k = -1.
Hence, the given system of equations will have no solution for k = -1.