1.

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

Answer»

Hello !!




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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