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Question

# Solve the following pair of equations: 5x−1+1y−2=2 6x−1−3y−2=1 (where x≠1,y≠2)

A

x=4, y=5

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B

x=5, y=4

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C

x=13,y=13

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D

None of the above

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Solution

## The correct option is A x=4, y=5 If we substitute 1x−1 as p and 1y−2 as q in the given equations. (As x≠1,y≠2) We get the equations as 5p+q=2 6q−3q=1 Now we can solve the pair of equation by method of elimination. 15p+3q=6 6p−3q=1 On adding both the above equations, we get p=13 Now by substituting the value in one of the equation we find q=13 As we have assumed p=1x−1 Therefore, p=1x−1=13 x=4 Similarly we assumed q=1y−2 Hence 1y−2=13 y−2=3 Thus x=4 and y=5

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