Answer:
Step-by-step explanation:
There is no solution for system of equations: [tex]a_1x+b_1y=c_1\\a_2y+b_2y=c_2[/tex]
if: [tex]a_1=a_2\quad and\quad b_1=b_2\quad and\quad \bold{c_1\ne c_2}[/tex]
so first, we we need to transform the equations to the form where the coefficients at y will be the same:
[tex]kx-3y=4\\ 4x-5y=7\\\\ (kx-3y)\cdot5=4\cdot5\\ (4x-5y)\cdot3=7\cdot3\\\\ 5kx-15y=20\\ 12x-15y=21[/tex]
Now we have b₁=b₂ and c₁≠c₂ so the system has no solution if a₁=a₂
5k = 12
÷5 ÷5
k = ¹²/₅