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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
x = -3 , y = -4
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Solution

## The correct option is A x=4, y=5If 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.....(i) 6q−3q=1.....(ii) Now we can solve the pair of equation by method of elimination.Multiply equation (i) by 3 15p+3q=6....(iii) Adding both (ii) and (iii) 21p=7 p=13 Now by substituting the value of p in (i), we get 5×13+q=2 ⇒q=2−53 ⇒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 or y=5 Thus x=4, y=5

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