The given pair of equations are: $$ 3x-2y=0\\ kx+5y=0 $$

Clearly,here, $a_1=3$, $a_2=k$, $b_1=-2$, $b_2=5$, $c_1=0$ and $c_2=0$.

Now, here, I have to find a value of $k$ for which the given system of equations has infinite solutions.

Hence, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

Thus, $\frac{3}{k} = -\frac{2}{5} = \frac{0}{0}$

But, $\frac{0}{0}$ is nonsense!

Why am I getting such a wrong result? Are $c_1$ and $c_2$ not equal to ZERO?

I will be thankful for help!

  • 3
    $\begingroup$ Here we see what happens when students learn the rules, but not the mathematics behind the rules: they forget the caveats and special cases. $c_1=c_2=0$ is a special case, and lets you skip $\frac{c_1}{c_2}$ altogether. Whatever theorem you're using should mention that $c_2\neq0$. $\endgroup$
    – Arthur
    Aug 27, 2017 at 7:01

3 Answers 3


Hint You can think this problem geometrically. A pair of equations has one solution if the lines that the equations represent intersect exactly once (the slopes of the lines differ from each other). Then of course for infinitely many solutions the lines must be exactly the same. What does this mean in your case?


First, if an equation has no unique solution,

$Δ = 0$,

$3×5 - (-2)k=0$


Using Gaussian Elimination, $ \left[\begin{array}{rr|r}3 & -2 & 0 \\k & 5 & 0 \\\end{array}\right]$~ $ \left[\begin{array}{rr|r}7.5 & -5 & 0 \\k & 5 & 0 \\\end{array}\right]$~$ \left[\begin{array}{rr|r}7.5 & -5 & 0 \\k + 7.5 & 0 & 0 \\\end{array}\right]$

Therefore, when there is infinite many solutions, $k+7.5=0$ must be true.

Then you have $k+7.5=0$ , i.e. $k=-7.5$

Let $x=t$, where $t$ is any real number, then the solutions of the equation are $(x,y)=$ $(t,\frac{3}{2}t)$.

  • $\begingroup$ Calculus ? I would say Linear algebra. $\endgroup$
    – user312097
    Aug 27, 2017 at 7:29
  • $\begingroup$ Yes, but then because it involves calculating infinite amount of solutions, matrixs would do much better. $\endgroup$ Aug 27, 2017 at 7:30
  • $\begingroup$ @Cynplytholowazy Yea but that doesn't mean it involves calculus. $\endgroup$
    – user441558
    Aug 27, 2017 at 7:52
  • $\begingroup$ Yea but then how about those involving 3, 4 or even more variables? That is definitely better using calculus. It's a good practice to use calculus, because in case you met a much more complicated ones you are still able to solve them much more quickly. $\endgroup$ Aug 27, 2017 at 7:55
  • $\begingroup$ @Cynplytholowazy I'm surely confused. It seems that Gaussian elimination is a technique in calculus, or is it? $\endgroup$
    – user441558
    Aug 27, 2017 at 7:57

It says that your rule, which you try to use is wrong.

The right rule is the following.

For $a_1^2+b_1^2\neq0$ and $a_2^2+b_2^2\neq0$ the system $$a_1x+b_1y=c_1$$ $$a_2x+b_2y=c_2$$ has infinitely many solutions if and only if $$a_1b_2-a_2b_1=c_1b_2-c_2b_1=a_1c_2-a_2c_1=0.$$

For your system it happens when the following equality holds. $$3\cdot5-k\cdot(-2)=0,$$ which gives, $k=-\frac{15}{2}$.

For this value your system it's $$3x-2y=0$$ and $$3x-2y=0,$$ which has infinitely many solutions: $$\left\{\left(t,\frac{3}{2}t\right)|t\in\mathbb R\right\}$$


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