Hello, WTF!
Here's another approach . . . Try to solve the systems.
1. For what value of \(\displaystyle n\) does .\(\displaystyle \begin{Bmatrix}2x\,+\,y\:=\:n\\x\,+\,2y\:=\:7\end{Bmatrix}\) .have infinitely many solutions?
A system has infinitely many solutions if we get a true statement, like: \(\displaystyle 0\,=\,0\) or \(\displaystyle 2\,=\,2\).
Multiply the second equation by -2:
. . . . \(\displaystyle 2x\,+\,4y\:=\:\,n\)
. . . .-\(\displaystyle 2x\,-\,4y\:=\:\)-\(\displaystyle 14\)
Add the equations:
.\(\displaystyle 0\:=\:n\,-\,14\)
We get a true statement "0 = 0" if \(\displaystyle n\,=\,14\).
2. For what value of \(\displaystyle n\) does .\(\displaystyle \begin{Bmatrix}4x\,-\,6y\:=\:5\\2x\,+\,y\:=\:2\end{Bmatrix}\) .have no solution?
A system has no solution if we get a false statement, like \(\displaystyle 1 = 2\)
. . or we get answers which are undefined.
Multiply the second equation by -2:
. . . . .\(\displaystyle 4x\,-\;6y\:=\;5\)
. . . .-\(\displaystyle 4x\,-\,2ny\:=\,\)-\(\displaystyle 4\)
Add the equations:
.-\(\displaystyle 6y\,-\,2ny\:=\:1\)
. . . Factor:
.-\(\displaystyle 2(3\,+\,n)y\:=\:1\:\:\Rightarrow\;\;y\,=\,-\frac{1}{2(n+3)\)
\(\displaystyle y\) has a value for every value of \(\displaystyle n\) . . . except \(\displaystyle n\,=\,\)-\(\displaystyle 3\)
Therefore, when \(\displaystyle n\,=\,\)-\(\displaystyle 3\), the system has no solution.
