K=5 which makes y=0
Plug in -16 for x which means y=7
Therefore, - 16,7 will be the value of x,y that has no solution.
If the system of linear equations above has no solution, and k is a constant, the value of k would be 3!
1. Let's get rid of the variable y
Multiply the first equation by 5 and the second by 3
Now subtract from equation(2)/equation(1) -> 12x-5kx=40
x(12-5k)=40 This equation is always solvable except if 12-5k=0, and this happens if k=12/5=2.4
There is no solution for system of equations:
so first, we we need to transform the equations to the form where the coefficients at y will be the same:
Now we have b₁=b₂ and c₁≠c₂ so the system has no solution if a₁=a₂
5k = 12
k = ¹²/₅
2x + 5y...