Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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..

share|improve this question
    
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
add comment

3 Answers

up vote 2 down vote accepted

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.

share|improve this answer
    
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 –  ZafarS Oct 6 '12 at 13:16
    
Nevermind I understand –  ZafarS Oct 6 '12 at 13:41
add comment

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

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

share|improve this answer
add comment

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 $x=y$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.