For what value ok K will …

The given equations are
{tex}2x - 3y - 7 = 0,{/tex}
{tex}(k +1) x + (1 - 2k) y + (4 - 5k) = 0.{/tex}
These equations are of the form
{tex}a_1x + b_1y+ c_1 = 0,\ a_2x + b_2y +c_2= 0,{/tex}
where {tex}a_1 = 2,\ b_1= -3,\ c_1= -7\ and\ a_2 =(k + 1),\ b_2= (1 - 2k),\ c_2= (4 - 5k){/tex}
Let the given system of equations have infinitely many solutions.
Then, {tex}\frac { a _ { 1 } } { a _ { 2 } } = \frac { b _ { 1 } } { b _ { 2 } } = \frac { c _ { 1 } } { c _ { 2 } }{/tex}
{tex}\Rightarrow \quad \frac { 2 } { ( k + 1 ) } = \frac { - 3 } { ( 1 - 2 k ) } = \frac { - 7 } { ( 4 - 5 k ) }{/tex}
{tex}\Rightarrow \frac { 2 } { ( k + 1 ) } = \frac { 3 } { ( 2 k - 1 ) } = \frac { 7 } { ( 5 k - 4 ) }{/tex}
{tex}\Rightarrow \quad \frac { 2 } { ( k + 1 ) } = \frac { 3 } { ( 2 k - 1 ) }{/tex} and {tex}\frac { 3 } { ( 2 k - 1 ) } = \frac { 7 } { ( 5 k - 4 ) }{/tex}
{tex}\Rightarrow{/tex} 4k - 2 = 3k + 3 and 15k -12 = 14k - 7
{tex}\Rightarrow{/tex} k = 5 and k = 5.
Hence, k = 5.