Given,
kx+10y=3 .......(1)
9x-15y=-19 .......(2)
Let two system of equations be
ax+by=c
px+qy=d
The above system of equations have no solution when
[tex]\frac{a}{p}=\frac{b}{q}=\ne\frac{c}{d}[/tex]Comapring with the given equations, we get
a=k, b=10, c=3 p=9, q=-15, d=-19.
So, the given system of equations has no solution if
[tex]\begin{gathered} \frac{k}{9}=\frac{10}{-15}\ne\frac{3}{-19} \\ \end{gathered}[/tex][tex]\begin{gathered} \frac{k}{9}=\frac{10}{-15} \\ k=\frac{10}{-15}\times9 \\ k=\frac{-2\times9}{3} \\ k=-6 \end{gathered}[/tex]So, the value of k is -6.
