Question

For how many values of k does the following system of equations have at least one solution?
\[x + y = 1;\,kx + y = 3;\,x + ky = 5\]
A) 0
B) 1
C) 2
D) Infinitely many

Answer 1

Hint: The question given here deals with three simultaneous equations but only in two variables. A multiplication factor $k$ is present in two of the equations. We need to find the values of $k$ for which three equations will have the same solutions $x$ and $y$.

Complete step by step solution:
Given equations,
\[x + y = 1;\,kx + y = 3;\,x + ky = 5\]
Solving
\[
  \;\;x + y = 1...(i) \\
  \,kx + y = 3...(ii) \\
  x + ky = 5...(iii) \\
 \]
Subtracting (i) from (ii) we get
\[
  kx + y = 3 \\
   - \,(x + y) = 1 \\
   \Rightarrow x = \dfrac{2}{{k - 1}} \\
 \]
Subtracting (i) from (iii)
\[
  x + ky = 5 \\
   - \,(x + y) = 1 \\
   \Rightarrow y = \dfrac{4}{{k - 1}} \\
 \]
Multiplying (ii) with k and and subtracting (iii) from it
\[
  {k^2}x + ky = 3k \\
   - \,(x + ky) = 5 \\
   \Rightarrow y = \dfrac{{3k - 5}}{{{k^2} - 1}} \\
 \]
Multiplying (iii) with k and and subtracting (ii) from it

\[
  {k^2}y + kx = 5k \\
   - \,(y + kx) = 3 \\
   \Rightarrow y = \dfrac{{5k - 3}}{{{k^2} - 1}} \\
 \]
The values of x and y for all the equations will be the same. So, taking ratio of x and y
$
   \Rightarrow \dfrac{{3k - 5}}{{5k - 3}} = \dfrac{2}{4} \\
   \Rightarrow 6k - 10 = 5k - 3 \\
   \Rightarrow k = 7 \\
 $
Therefore, the equations will have roots for exactly one value of k. So, Option (B) is correct.

Note:
While solving the simultaneous equations keep in mind.
> No. of variables in the equation.
> If the number of equations is greater than or equal to the number of variables it is possible to find a solution.
> All the simultaneous equations to be solved will have the same roots.