if the system of equation 6 X + 2 Y is equal to 3 and K X + Y is equal to 2 has a unique solution find the value of k
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6X + 2Y = 3
6X + 2Y - 3 = 0 ------------(1)
And,
KX + Y = 2
KX + Y - 2 = 0---------(1)
These equations are of the form A1X + B1Y + C1 = 0 and A2X + B2Y + C2 = 0.
Where,
A1 = 6 , B1 = 2 and C1 = -3
A2 = K , B2 = 1 and C2 = -2
Therefore,
A1/A2 = 6/K , B1/B2 = 2/1 and C1/C2 = 3/2
For unique solution, we must have
A1/A2 # B1/B2 [ Where # stands for not equal]
6/K # 2
2k # 6
k # 6/2
K # 3
Therefore,
K has all real values , other than 3.
6X + 2Y = 3
6X + 2Y - 3 = 0 ------------(1)
And,
KX + Y = 2
KX + Y - 2 = 0---------(1)
These equations are of the form A1X + B1Y + C1 = 0 and A2X + B2Y + C2 = 0.
Where,
A1 = 6 , B1 = 2 and C1 = -3
A2 = K , B2 = 1 and C2 = -2
Therefore,
A1/A2 = 6/K , B1/B2 = 2/1 and C1/C2 = 3/2
For unique solution, we must have
A1/A2 # B1/B2 [ Where # stands for not equal]
6/K # 2
2k # 6
k # 6/2
K # 3
Therefore,
K has all real values , other than 3.
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