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Find the value of k for which each of the following system of equations have no solution :2 x+k y=115 x 7 y=5
The given system of equation is
2x + ky - 11 =0
5x − 7y - 5 = 0
The system of equation is of the form
`a_1x + b_1y + c_1 = 0`
`a_2x + b_2y + c_2 = 0`
Where, `a_1 = 2,b_1 = k, c_1 = -11`
And `a_2 = 5, b_2 = -7, c_2 = -5`
For a unique solution, we must have
`a_1/a_2 - b_1/b_2 != c_1/c_2`
`=> 2/5 = k/(-7) != (-11)/(-5)`
Now
`2/5 = k/(-7)`
`=> 2 xx (-7) = 5k`
`=> 5k = -14`
`=> k = (-14)/5`
Cleary for `(-14)/5` we have `k/(-7) != (-11)/(-5)`
Hence, the given system of equation will have no solution if `k = (-14)/5`
Page 2
Find the value of k for which the following system of equations has a unique solution:
kx + 3y = 3
12x + ky = 6
kx + 3y = 3
12x + ky = 6
For no solution `a_1/a_2 - b_1/b_2 != c_1/c_2`
`=> k/12 = 2/k != 3/6`
`k/12 = 3/k`
`k^2 = 36`
`k = +- 6` i.e k = 6, -6
Also
`3/k != 3/6`
`(3 xx 6)/3 != k`
`k != 6`
k = -6 satisfies both the condition
Hence k = -6
Concept: Pair of Linear Equations in Two Variables
Is there an error in this question or solution?
Page 3
The given system of equation may be written as
4x + 6y - 11 = 0
2x + ky - 7 = 0
The system of equation is of the form
`a_1x + b_1y + c_1 = 0`
`a_2x + b_2y + c_2 = 0`
Where `a_1 = 4, b_1 = 6, c_1 = -11`
And `a_2 = 2, b_2 = k, c_2 = -7`
For a unique solution, we must have
`a_1/a_2 = b_1/b_2 != c_1/c_2`
Now
`a_1/a_2 = b_1/b_2`
`=> 4/2 = 6/k`
`=> 4k = 12`
`=> k = 12/4 = 3`
Clearly, for this value of k, we have
`a_1/a_2 = b_1/b_2 != c_1/c_2
Hence, the given system of equation is inconsistent, when k = 3
Page 4
The given system of the equation may be written as
αx + 3y -α - 3 = 0
12x + αy - α = 0
The system of equation is of the form
`a_1x + b_1y + c_1 = 0`
`a_2x + b_2y + c_2 = 0`
Where `a_1 = α, b_1 = 3, c_1 = -(α - 3)`
And `a_2 = 12, b_2 = α, c_2 = -α`
For a unique solution, we must have
`a_1/a_2 - b_1/b_2 != c_1/c_2`
`=> α/12 = 3/α != (-(α - 3))/(-α)`
Now,
`3/α != (-(α - 3))/(-α)`
`=> 3/α != (α - 3)/α`
`=> 3 != α - 3`
`=> 3 = 3 != α`
`=> 6 != α`
`=> α != 6`
And
`α/12 = 3/α`
`=> α^2 = 36`
`=> α +- 6`
`=> α = -6` [∵ `α != 6`]
Hence, the given system of equation will have no solution if α = -6