B
C
A
D
3x+y=3 x-y=5
3x-y=3 x-y=5
2x+y=3 x+y=5
3x+y=3 x+y=5
By Team Aakash Byju's | 9th November 2022
Let us consider each given option and substitute the values of x and y to check if they satisfy the equation or not.
Substitute x=2 and y=-3 in the given options A,B,C and D.
For A, equations are
3x+y=3 and x+y=5
Now substitute x=2 and y=-3 in the above equations.
3x+y=3 => 3(2)+(-3) = 3
=> 6 - 3 = 3 so, 3 = 3
x+y=5
=> -1 = 5 which is not true.
Hence, option A does not have the given unique solution.
=> 2 + (-3) = 5 => 2 - 3 = 5
For B, equations are
2x+y=3 and x+y=5
Now substitute x=2 and y=-3 in the above equations.
2x+y=3 => 2(2)+(-3) = 3
=> 4 - 3 = 3 => 1 = 3 which is not true.
x+y=5 => 2 + (-3) = 5
=> 2 - 3 = 5 => -1 = 5 which is not true.
Hence, option B does not have the given unique solution.
For C, equations are
3x-y=3 and x-y=5
Now substitute x=2 and y=-3 in the above equations.
3x-y=3 => 3(2)-(-3) = 3
=> 6 + 3 = 3 => 9 =3
x-y=5 => (2) - (-3) = 5
=> 2 + 3 = 5 => 5 = 5 which is not true.
Hence, option C does not have the given unique solution.
For D, equations are
3x+y=3 and x-y=5
Now substitute x=2 and y=-3 in the above equations.
3x+y=3 => 3(2)+(-3) = 3
=> 6 - 3 = 3 => 3 = 3
x-y=5 => (2) - (-3) = 5
=> 2 + 3 = 5 => 5 = 5 Both are true.
Hence, option D has the given unique solution.