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# 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) 0B) 1C) 2D) Infinitely many

Last updated date: 20th Jul 2024
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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.