For what values of b does the system have no solutions? Help me please with system.
It turned out this way:
Multiply system 2
4x + by = 4 
2bx + 2y = 4
The system has no solutions if:
4x + by ≠ 2bx + 2y
x (4 - 2b) ≠ y (2 - b)
2x (2 - b) ≠ y (2 - b)
For any b ≠ 2, the system will have only one solution.
And when b = 2 - infinite number of solutions:
4x + 2y = 4
2x + y = 2
Does it mean that for any b there will be such that the system has no solution?
It seems to me that when b = -2 will not solutions ....
The system
 A: You have the following system with $b\not = 0$:
$$4x+by=4\Rightarrow y=-\frac{4}{b}x+\frac{4}{b}$$
$$2bx+2y=4\Rightarrow y=-bx+2 $$
These equations are the equations of two straight lines with slopes $λ_1=-\frac{4}{b}$ and $λ_2=-b$.
These two straight lines have no solution when they do not intersect, meaning that they are parallel. Hence must:
$$λ_1=λ_2\Rightarrow -\frac{4}{b}=-b\Rightarrow b^2=4\Rightarrow b=-2 \ or \ b=2$$
When $b=2$ the two equations have infinite solutions, hence it's rejected.
So $b=-2$.
If $b=0$ then $x=2 \ and \ y=2$.
So after all the only solution to the problem is $b=-2$
A: Yes it is correct indeed the system is


*

*$4x+by=4$

*$bx+y=2$


and for $b=-2$


*

*$4x-2y=4\implies 2x-y=2$

*$-2x+y=2\implies 2x-y=-2$


We can solve the problem in a more sistematic way by the RREF on the augmented matrix, that is
$$\left[ {\begin{array}{rr|r}
4&b&4\\
b&1&2\\
\end{array} } \right]\to \left[ {\begin{array}{rr|r}
4b&b^2&4b\\
4b&4&8\\
\end{array} } \right]\to \left[ {\begin{array}{rr|r}
4b&b^2&4b\\
0&4-b^2&8-4b\\
\end{array} } \right]$$
and thus the system has no solution for


*

*$4-b^2=0\implies b=\pm 2$

*$8-4b\neq 0\implies b\neq2$


that is precisely $b=-2$.
