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Dear Student,
Q2) Quadratic equation: 5x+3y= 5
Option(A). (1,0)
Put x= 1 and y= 0 in the above equation.
5(1)+3(0)= 5+0=5
As (1,0) satisfies the above equation, it is a root of the given equation.
Option(B): (2,1)
Put x= 2 and y= 1 in the given equation.
5(2)+3(1)= 10+3=13 which is not equal to 5.
(2,1) does not satisfy the above equation. (2,1) is not a root of the given equation.
Option (C):(-1,-2)
Put x= -1 and y= -2in the given equation.
5(-1)+3(-2)= -5-6=-11 which is not equal to 5. Therefore, (-1,-2) is not a root of the given equation.
Option (D): (0,1)
Put x= 0 and y= 1 in the given equation.
5(0)+3(1)= 3 which is not equal to 5. Therefore, (0,1) is not a root of the given equation.
Therefore, Option (A) (1,0) is only the root of 5x+3y=5 amongst the given options.
Hence, Option (A) is the correct answer.
Kindly ask the remaining questions again in a separate thread so that we may provide you with some meaningful help. Looking forward to hear from you soon:)
Regards
Q2) Quadratic equation: 5x+3y= 5
Option(A). (1,0)
Put x= 1 and y= 0 in the above equation.
5(1)+3(0)= 5+0=5
As (1,0) satisfies the above equation, it is a root of the given equation.
Option(B): (2,1)
Put x= 2 and y= 1 in the given equation.
5(2)+3(1)= 10+3=13 which is not equal to 5.
(2,1) does not satisfy the above equation. (2,1) is not a root of the given equation.
Option (C):(-1,-2)
Put x= -1 and y= -2in the given equation.
5(-1)+3(-2)= -5-6=-11 which is not equal to 5. Therefore, (-1,-2) is not a root of the given equation.
Option (D): (0,1)
Put x= 0 and y= 1 in the given equation.
5(0)+3(1)= 3 which is not equal to 5. Therefore, (0,1) is not a root of the given equation.
Therefore, Option (A) (1,0) is only the root of 5x+3y=5 amongst the given options.
Hence, Option (A) is the correct answer.
Kindly ask the remaining questions again in a separate thread so that we may provide you with some meaningful help. Looking forward to hear from you soon:)
Regards
