Why am i getting an incorrect result (0/0)? 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!
 A: First, if an equation has no unique solution,
$Δ = 0$,
$3×5 - (-2)k=0$ 
$2k=15$
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  $(x,y)=$    $(t,\frac{3}{2}t)$.
A: 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? 
A: 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:
$$\left\{\left(t,\frac{3}{2}t\right)|t\in\mathbb R\right\}$$
