Number of solutions of a linear system --- dependence on parameters For which values of $a$ and $b$ does the system:
$3x+ay=b$
$ax+3y=b$
have:
$a)$ no solutions,
$b)$ 1 solution,
$c)$ infinite solutions.
How to tackle this? Please don't use advanced mathematics..
 A: If $(x,y)$ is a solution, then by subtracting we get 
$$3x-ax+ay-3y=b-b=0.$$
This can be rewritten as 
$$(3-a)(x-y)=0.$$
If $a=3$, the equation holds for all $x$ and $y$. So let's check what happens with our original equations when $a=3$.
We get $3x+3y=b$, $3y+3x=b$. For any $b$, there are infinitely many solutions, since we can arrange in infinitely many ways to have $x+y=b/3$.
Thus if $a=3$ and $b$ is anything, there is a unique solution.
We have dealt with the case $a=3$. Now suppose $a\ne 3$. Then from the equation $(3-a)(x-y)=0$ we get $x=y$.
Now go back to the original equations, with $a\ne 3$ and $x=y=t$.
The first equation reads $3t+at=b$, the second reads $at+3t=b$, the same. If $a+3=0$ and $b\ne 0$, there is no solution. So there is no solution if $a=-3$ and $b\ne 0$. 
If $a=-3$, and $b=0$, there are infinitely many solutions, since any $t$ works.
If $a\ne -3$, we can comfortably divide, getting the unique soluton $x=y=t=\dfrac{b}{a+3}$.
Summary: (1) No solution if $a=-3$ and $b\ne0$.
(2) Infinitely many solutions if $a=3$ and $b$ is anything; also if $a=-3$ and $b=0$.
(3) Otherwise, unique solution.
A: $$\begin{cases}3x+ay=b\\ax+3y=b\end{cases}\leftrightarrow
\begin{cases}3x+ay=b\\(3-\frac{a^{2}}{3})y=b(1-\frac{a}{3})\end{cases}$$
a.no solutions: $3-\frac{a^{2}}{3}=0$ and $b(1-\frac{a}{3})\neq0$, $a=-3$ and $b\neq0$;
b.1 solution:  $3-\frac{a^{2}}{3}\neq0$, $a\neq3$ and $a\neq-3$;
c.infinite solution: $3-\frac{a^{2}}{3}=0$ and $b(1-\frac{a}{3})=0$, $a=3$, $b$ arbitrary or $a=-3$, $b=0$.
A: From the first equation, you will get $x=\frac{b-ay}{3}$. Then plug it to the second equation, $$a\frac{b-ay}{3}+3y=b$$
$$\left(3-\frac{a^2}{3}\right)y=b-\frac{ab}{3}$$
So if $a\neq \pm 3$, $y=\left(b-\frac{ab}{3}\right)/\left(3-\frac{a^2}{3}\right)$ and you can get $x$.
If $a=3$, the original equations become $x+y=\frac{b}{3}$. There are infinitely many sets of solutions.
If $a=-3$ and $b\neq 0$, no solution.
If $a=-3$ and $b=0$, solutions are any $x=y$.
