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  • A pair of linear equations which has a unique solution x = 2, y = –3 is (A) x + y = –1 ; 2x – 3y = –5 (B) 2x + 5y = –11 ; 4x + 10y = –22 (C) 2x – y = 1 ; 3x + 2y = 0 (D) x– 4y –14 = 0 ; 5x – y – 13 = 0

A pair of linear equations which has a unique solution x = 2, y = –3 is
(A) x + y = –1 ; 2x – 3y = –5
(B) 2x + 5y = –11 ;  4x + 10y = –22
(C) 2x – y = 1 ; 3x + 2y = 0
(D) x– 4y –14 = 0 ; 5x – y – 13 = 0

Answers (1)

Answer: B, D
Solution:
A pair which have unique solution x = 2, y = –3 will satisfy L.H.S. = R.H.S. if we put x = 2 and y = –3
Put x = 2, y = –3 in option (A)
x + y = –1        ;           2x – 3y = –5
2 – 3 = –1        ;           2(2) – 3 (–3) = –5
–1 = –1 (True) ;           4 + 9 = –5
   13 = – 5 (False)
Put x = 2, y = –3 in option (B)
2x + 5y = –11  ;           4x + 10y = –22
2(2) + 5(–3) = –11  ;    4(2) + 10 (–3) = –22
4 – 15 = –11    ;           8 – 30 = –22
–11 = –11 (True)         –22 = –22 (True)
Put x = 2, y = – 3 in option (C)
2x – y = 1        ;           3x + 2y = 0
2(2) – (–3) = 1 ;           3(2) + 3 (–3) = 0
4 + 3 = 1          ;           6 – 6 = 0
7 = 1 (False)    ;           0 = 0 (True)
Put x = 2, y = – 3 in option (D)
x – 4y – 14 = 0            ;           5x – y – 13 = 0
2 – 4 (–3) – 14 = 0      ;           5(2) – (–3) – 13 = 0
2 + 12 – 14 = 0            ;           10 + 3 – 13 = 0
0 = 0 (True)                 ;           13 – 13 = 0
      0 = 0 (True)
By putting the value of x = 2, y = –3 in A, B, C, D; two options B, D satisfy L.H.S = R.H.S
Hence both (B, D) are correct

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