Question

Find the value of k for which the system of equations $kx-y=2$ , $6x-2y=3$ has a unique solution?

Answer 1

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

Complete step-by-step answer:
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 a 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 equations is analogous to the planes
Here we have 2 variables, so in our case:
Our system of equations is analogous to 2 straight lines.
We have 3 possibilities
(a) No solutions (b) Infinite solutions (c) 1 solution
(a) No solution:
If 2 straight lines(infinitely long) have 0 solutions then they must not intersect anywhere that means they are parallel lines.
For 2 lines to be parallel their x-coordinates and y-coordinates must be proportional but constants must not be in proportion to them.
In mathematical way:
If system of equations are $ax+by+c=0$ , $dx+ey+f=0$ then
$\dfrac{a}{d}=\dfrac{b}{e}\ne \dfrac{c}{f}$
$\Rightarrow$ No solutions
(b) Infinite solutions:
If 2 infinitely long straight lines have infinite solutions, then they must be coinciding lines, as infinite intersection points implies infinite solutions their x=coordinate, y-coordinate and constants must be in proportion.
In mathematical way:
If system of equations are $ax+by+c=0$ , $dx+ey+f=0$ then
$\dfrac{a}{d}=\dfrac{b}{e}=\dfrac{c}{f}$
$\Rightarrow$ Infinite solutions.
(c) 1 solution:
If 2 infinitely long straight lines have 1 solution. They must be intersecting at only 1 point.
$\Rightarrow$ If not the above 2 cases then the system of equations satisfy this case.
The system of equation given in question:
$kx-y=2$
$6x-2y=3$
From previous conditions we can say:
$a=k,b=-1,c=-2$
$d=6,e=-2,f=-3$
As condition for 1 solution:
$\dfrac{a}{d}\ne \dfrac{b}{e}$
By substituting a,b,c,d,e,f we get
$\dfrac{k}{6}\ne \dfrac{1}{2}$
By solving, we get
$k\ne \dfrac{6}{2}\text{ }\Rightarrow \text{k}\ne \text{3}$
By above we can say that all the real numbers except 3 we can say that only unique solutions are possible.

Note: The idea of projecting the system of equations analogous to straight lines is crucial. Remember whenever you see 2 variables equations just project the idea of it as straight lines. Be careful while deriving the conditions from straight line equations given. The constants represent the distance between straight lines. Thus they must not be equal or else they will become coincident lines. Generally students confuse and forget to take the condition that their constants must not be in proportion but it is the important part of the solution.