# 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..

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This system can never have infinite solutions. The solutions are always (pairs of) finite numbers. You mean infinitely many solutions. –  Chris Eagle Oct 4 '12 at 18:18

## 3 Answers

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.

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How about $a=3$? There are also infinitely many solutions! –  Alfred Chern Oct 4 '12 at 18:04
@AlfredChern: Thanks! Somehow I had written $3x+3x$ instead of $3x+3y$. –  André Nicolas Oct 4 '12 at 18:15
But how can b be 0 at all? $\dfrac{3}{a} = \dfrac{a}{3} = \dfrac{b}{b}$ If b is 0, then $\dfrac{0}{0}$ which is impossible –  JohnPhteven Oct 6 '12 at 13:16
Nevermind I understand –  JohnPhteven Oct 6 '12 at 13:41

$$\begin{cases}3x+ay=b\\ax+3y=b\end{cases}\leftrightarrow \begin{cases}3x+ay=b\\(3-\frac{a^{2}}{3})y=b(1-\frac{a}{3})\end{cases}$$

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 $a=-3$, $b=0$.

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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 $x=y$.

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