Value of k for which the system is inconsistent Determine the   value   of  k for which   the system  is  inconsistent,   if  no  such    value   exists  state   that.
$\begin{cases}5x + 7y = -15\\
–x + ky = 3\end{cases}$
Solution: No    such    k exists.
Why is there no k? How can you determine this?
Thank you.
 A: Consider the system of equations $$\begin{cases}a_1x + b_1y = c_1\\
a_2x + b_2y = c_2\end{cases}\tag1$$
It has a unique solution if and only if $~\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2}~$.
It has an infinitely many solutions if and only if $~\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}~$.

A system of equations is said to be consistent if it has at least one solution.

In the above two cases the system is consistent.
It has no solution if and only if $~\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}~$.
In the above case the system is inconsistent.


Here the system of equation is $$\begin{cases}5x + 7y = -15\\
–x + ky = 3\end{cases}\tag 2$$
The system is inconsistent, if 
$~\dfrac{5}{-1}=\dfrac{7}{k}\ne\dfrac{-15}{3}~\implies ~-5=\dfrac{7}{k}\ne -5~$
There is no $~k~$ for which the above condition is satisfied. 
Hence there is no value of k for which the system is inconsistent.
A: To say that the system is inconsistent is to say that there are no values of $x$ and $y$ that satisfy it.  If we add $5$ times the second equation to the first, we get $$(7+5k)y = 0$$  No matter what the value of $k$ is, we can take $y=0$ to satisfy the equation.  Substituting into the first equation, we find $x=-3$.  Now we can check that $x=-3, y=0$ satisfy both equations, no matter what the value of $k$ is.  That is, no value of $k$ makes the system inconsistent. 
