determine the value of k for which the pair of linear equations 2 X + 3 Y - 5 = 0, kx - 6y- 8 = 20 has a unique solution
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Hii friend,
2X + 3Y -5 = 0......(1)
And,
KX -6Y -8 =20
KX -6Y -8-20 = 0
KX -6Y -28 = 0
These equations are in the form of AX1+BY1+C1 = 0 and AX2+BY2+C2 = 0
A1 = 2 , B1 = 3 and C1 = -5
And,
A2= K , B2 = -6 , C2 = -28
For a unique solution we must have,
A1/A2 is not equal to B1/B2
=> 2/K isn't equal 3/-6
=> 3×K = 2 × -6
=> 3K = -12
=> K = -12/3
=> K = -4.
HOPE IT WILL HELP YOU... :-)
2X + 3Y -5 = 0......(1)
And,
KX -6Y -8 =20
KX -6Y -8-20 = 0
KX -6Y -28 = 0
These equations are in the form of AX1+BY1+C1 = 0 and AX2+BY2+C2 = 0
A1 = 2 , B1 = 3 and C1 = -5
And,
A2= K , B2 = -6 , C2 = -28
For a unique solution we must have,
A1/A2 is not equal to B1/B2
=> 2/K isn't equal 3/-6
=> 3×K = 2 × -6
=> 3K = -12
=> K = -12/3
=> K = -4.
HOPE IT WILL HELP YOU... :-)
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