Question

Find the value of k for which the system of equations
$\begin{align}
  & kx-y=2 \\
 & 6x-2y=3 \\
\end{align}$
has no solution. Is there a value of k for which the system has infinitely many solutions?

Answer 1

Hint: For system of equations the solutions follow some conditions
If there are system of equation, namely $ax+by+c=0$ and $dx+ey+f=0$
Then, $\begin{align}
  & \dfrac{a}{d}=\dfrac{b}{e}=\dfrac{c}{f}\Rightarrow \text{Infinite solutions} \\
 & \dfrac{a}{d}=\dfrac{b}{e}\ne \dfrac{c}{f}\Rightarrow \text{No solutions} \\
\end{align}$

Complete step-by-step solution-
Definition of system of equations:
If simultaneously we have more than one equation, then the set of those equations is called a system of equations. We can project systems of equations as lines, planes etc. depending on number of variables.
If we have 2 variables:
Then the system of equations is analogous to straight lines.
If we have 3 variables:
Then system of equation is analogous to the planes
Here we have two variables. So, in our case:
Our system of equations is analogous to straight lines
We have 3 possibilities
(a) No solutions
(b) Infinite solutions
(c) One solution

(a) No solutions:
If two straight lines (infinitely long) have 0 solutions then they must not intersect anywhere that means they are parallel lines.
For two lines to be parallel their x - coordinates and y – coordinates must be proportional but constant must not be in proportion to them.
In mathematical way:
If system of equations are
$ax+by+c=0$ and $dx+ey+f=0$
then $\dfrac{a}{d}=\dfrac{b}{e}\ne \dfrac{c}{f}$
$\Rightarrow $No solutions
(b) Infinite solutions:
If two infinitely long straight lines have infinite solutions then they must be coincident lines, as infinite intersection point implies in finite solutions their x – coordinates and y – coordinates and constants must be in proportion.
In mathematical way:
 If system of equations are
$\begin{align}
  & ax+by+c=0 \\
 & dx+ey+f=0 \\
\end{align}$
Then $\dfrac{a}{d}=\dfrac{b}{e}=\dfrac{c}{f}$
$\Rightarrow $ Infinite solutions
(c) One solution
If two infinitely long straight lines have one solution, they must be intersecting at only one point
$\Rightarrow $ If not the above 2 cases then the system of equations satisfy this case.
The system of equation given in question:
$\begin{align}
  & kx-y=2 \\
 & 6x-2y=3 \\
\end{align}$
From previous conditions we can say
$\begin{align}
  & a=k,b=-1,c=-2 \\
 & d=6,e=-2,f=-3 \\
\end{align}$
As condition for no solution:
$\dfrac{a}{d}=\dfrac{b}{e}\ne \dfrac{c}{f}$
By substituting a, b, c, d, e, f we get:
$\dfrac{k}{6}=\dfrac{1}{2}\ne \dfrac{2}{3}$
By solving we get:
$k=\dfrac{6}{2}\Rightarrow k=3$
As $\dfrac{b}{e}\ne \dfrac{c}{f}$
$\Rightarrow $ Infinitely many solutions is not possible
Hence, for k = 3 system has no solution. There does not exist any k for which system has infinitely many solutions.

Note: The idea of projecting a system analogous to a straight line is crucial. Remember whenever you see two variable equations just project it as a straight line.