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Question

The values of x and y satisfying the two equations 32x+33y=31, 33x+32y=34 respectively will be

A
1,2
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B
2,1
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C
0,0
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D
2,3
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Solution

The correct option is B 2,1
Multiplying 32x+33y=31 with 33, we get 1056x+1089y=1023 ---- (1)

Multiplying 33x+32y=34 with 32, we get 1056x+1024y=1088 ---- (2)
Subtracting (2) from (1),
1056x+1089y1056x1024y=10231088
65y=65
y=1

Substituting y=1 in 32x+33y=31 , we get $ 32x - 33 = 31
32x=64
x=2

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